【原创】AC自动机小结

有了KMP和Trie的基础，就可以学习神奇的AC自动机了。AC自动机其实就是在Trie树上实现KMP，可以完成多模式串的匹配。

AC自动机其实就是创建了一个状态的转移图，思想很重要。

http://blog.csdn.net/niushuai666/article/details/7002823

http://hi.baidu.com/nialv7/item/ce1ce015d44a6ba7feded52d

AC自动机专题训练链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=25605#overview 这里我提交的代码是公开的，可以看到

题目来源：http://www.notonlysuccess.com/index.php/aho-corasick-automaton/

写AC自动机的代码风格是向昀神学的，好简洁，写起来好棒的感觉。

1、HDU 2222 Keywords Search 最基本的入门题了

就是求目标串中出现了几个模式串。

很基础了。使用一个int型的end数组记录，查询一次。

//======================
// HDU 2222
// 求目标串中出现了几个模式串
//====================
#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
#include <queue>
using namespace std;

struct Trie
{
    int next[500010][26],fail[500010],end[500010];
    int root,L;
    int newnode()
    {
        for(int i = 0;i < 26;i++)
            next[L][i] = -1;
        end[L++] = 0;
        return L-1;
    }
    void init()
    {
        L = 0;
        root = newnode();
    }
    void insert(char buf[])
    {
        int len = strlen(buf);
        int now = root;
        for(int i = 0;i < len;i++)
        {
            if(next[now][buf[i]-'a'] == -1)
                next[now][buf[i]-'a'] = newnode();
            now = next[now][buf[i]-'a'];
        }
        end[now]++;
    }
    void build()
    {
        queue<int>Q;
        fail[root] = root;
        for(int i = 0;i < 26;i++)
            if(next[root][i] == -1)
                next[root][i] = root;
            else
            {
                fail[next[root][i]] = root;
                Q.push(next[root][i]);
            }
        while( !Q.empty() )
        {
            int now = Q.front();
            Q.pop();
            for(int i = 0;i < 26;i++)
                if(next[now][i] == -1)
                    next[now][i] = next[fail[now]][i];
                else
                {
                    fail[next[now][i]]=next[fail[now]][i];
                    Q.push(next[now][i]);
                }
        }
    }
    int query(char buf[])
    {
        int len = strlen(buf);
        int now = root;
        int res = 0;
        for(int i = 0;i < len;i++)
        {
            now = next[now][buf[i]-'a'];
            int temp = now;
            while( temp != root )
            {
                res += end[temp];
                end[temp] = 0;
                temp = fail[temp];
            }
        }
        return res;
    }
    void debug()
    {
        for(int i = 0;i < L;i++)
        {
            printf("id = %3d,fail = %3d,end = %3d,chi = [",i,fail[i],end[i]);
            for(int j = 0;j < 26;j++)
                printf("%2d",next[i][j]);
            printf("]\n");
        }
    }
};
char buf[1000010];
Trie ac;
int main()
{
    int T;
    int n;
    scanf("%d",&T);
    while( T-- )
    {
        scanf("%d",&n);
        ac.init();
        for(int i = 0;i < n;i++)
        {
            scanf("%s",buf);
            ac.insert(buf);
        }
        ac.build();
        scanf("%s",buf);
        printf("%d\n",ac.query(buf));
    }
    return 0;
}

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2、HDU 2896 病毒侵袭

这题和上题差不多，要输出出现模式串的id,用end记录id就可以了。还有trie树的分支是128的

题解here

//============================================================================
// Name        : HDU.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;

struct Trie
{
    int next[210*500][128],fail[210*500],end[210*500];
    int root,L;
    int newnode()
    {
        for(int i = 0;i < 128;i++)
            next[L][i] = -1;
        end[L++] = -1;
        return L-1;
    }
    void init()
    {
        L = 0;
        root = newnode();
    }
    void insert(char s[],int id)
    {
        int len = strlen(s);
        int now = root;
        for(int i = 0;i < len;i++)
        {
            if(next[now][s[i]] == -1)
                next[now][s[i]] = newnode();
            now=next[now][s[i]];
        }
        end[now]=id;
    }
    void build()
    {
        queue<int>Q;
        fail[root] = root;
        for(int i = 0;i < 128;i++)
            if(next[root][i] == -1)
                next[root][i] = root;
            else
            {
                fail[next[root][i]] = root;
                Q.push(next[root][i]);
            }
        while(!Q.empty())
        {
            int now = Q.front();
            Q.pop();
            for(int i = 0;i < 128;i++)
                if(next[now][i] == -1)
                    next[now][i] = next[fail[now]][i];
                else
                {
                    fail[next[now][i]] = next[fail[now]][i];
                    Q.push(next[now][i]);
                }
        }
    }
    bool used[510];
    bool query(char buf[],int n,int id)
    {
        int len = strlen(buf);
        int now = root;
        memset(used,false,sizeof(used));
        bool flag = false;
        for(int i = 0;i < len;i++)
        {
            now = next[now][buf[i]];
            int temp = now;
            while(temp != root)
            {
                if(end[temp] != -1)
                {
                    used[end[temp]] = true;
                    flag = true;
                }
                temp = fail[temp];
            }
        }
        if(!flag)return false;
        printf("web %d:",id);
        for(int i = 1;i <= n;i++)
            if(used[i])
                printf(" %d",i);
        printf("\n");
        return true;
    }
};
char buf[10010];
Trie ac;
int main()
{
    int n,m;
    while(scanf("%d",&n) != EOF)
    {
        ac.init();
        for(int i = 1;i <= n;i++)
        {
            scanf("%s",buf);
            ac.insert(buf,i);
        }
        ac.build();
        int ans = 0;
        scanf("%d",&m);
        for(int i = 1;i <= m;i++)
        {
            scanf("%s",buf);
            if(ac.query(buf,n,i))
                ans++;
        }
        printf("total: %d\n",ans);
    }
    return 0;
}

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3、HDU 3065 病毒侵袭持续中

这题的变化也不大，就是需要输出每个模式串出现的次数，查询的时候使用一个数组进行记录就可以了

//============================================================================
// Name        : HDU.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
using namespace std;

char str[1010][100];
struct Trie
{
    int next[1010*50][128],fail[1010*50],end[1010*50];
    int root,L;
    int newnode()
    {
        for(int i = 0;i < 128;i++)
            next[L][i] = -1;
        end[L++] = -1;
        return L-1;
    }
    void init()
    {
        L = 0;
        root = newnode();
    }
    void insert(char s[],int id)
    {
        int len = strlen(s);
        int now = root;
        for(int i = 0;i < len;i++)
        {
            if(next[now][s[i]] == -1)
                next[now][s[i]] = newnode();
            now = next[now][s[i]];
        }
        end[now] = id;
    }
    void build()
    {
        queue<int>Q;
        fail[root] = root;
        for(int i = 0;i < 128;i++)
            if(next[root][i] == -1)
                next[root][i] = root;
            else
            {
                fail[next[root][i]] = root;
                Q.push(next[root][i]);
            }
        while(!Q.empty())
        {
            int now = Q.front();
            Q.pop();
            for(int i = 0;i < 128;i++)
                if(next[now][i] == -1)
                    next[now][i]=next[fail[now]][i];
                else
                {
                    fail[next[now][i]]=next[fail[now]][i];
                    Q.push(next[now][i]);
                }
        }
    }
    int num[1010];
    void query(char buf[],int n)
    {
        for(int i = 0;i < n;i++)
            num[i] = 0;
        int len=strlen(buf);
        int now=root;
        for(int i=0;i<len;i++)
        {
            now=next[now][buf[i]];
            int temp = now;
            while( temp != root )
            {
                if(end[temp] != -1)
                    num[end[temp]]++;
                temp = fail[temp];
            }
        }
        for(int i = 0;i < n;i++)
            if(num[i] > 0)
                printf("%s: %d\n",str[i],num[i]);
    }

};

char buf[2000010];
Trie ac;
void debug()
{
    for (int i = 0; i < ac.L; i++)
    {
        printf("id = %3d ,fail = %3d ,end = %3d, chi = [",i,ac.fail[i],ac.end[i]);
        for (int j = 0; j < 128; j++)
            printf("%2d ",ac.next[i][j]);
        printf("]\n");
    }
}
int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    int n;
    while(scanf("%d",&n) == 1)
    {
        ac.init();
        for(int i = 0;i < n;i++)
        {
            scanf("%s",str[i]);
            ac.insert(str[i],i);
        }
        ac.build();
        scanf("%s",buf);
        ac.query(buf,n);
    }
    return 0;
}

View Code

4、ZOJ 3430 Detect the Virus

主要是解码过程，解码以后就是模板题了。

求出现的模式串的种类数

分支需要256个

//============================================================================
// Name        : ZOJ.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
using namespace std;

struct Trie
{
    int next[520*64][256],fail[520*64],end[520*64];
    int root,L;
    int newnode()
    {
        for(int i = 0;i < 256;i++)
            next[L][i] = -1;
        end[L++] = -1;
        return L-1;
    }
    void init()
    {
        L = 0;
        root = newnode();
    }
    void insert(unsigned char buf[],int len,int id)
    {
        int now = root;
        for(int i = 0;i < len;i++)
        {
            if(next[now][buf[i]] == -1)
                next[now][buf[i]] = newnode();
            now = next[now][buf[i]];
        }
        end[now] = id;
    }
    void build()
    {
        queue<int>Q;
        fail[root] = root;
        for(int i = 0;i < 256;i++)
            if(next[root][i] == -1)
                next[root][i] = root;
            else
            {
                fail[next[root][i]]=root;
                Q.push(next[root][i]);
            }
        while(!Q.empty())
        {
            int now = Q.front();
            Q.pop();
            for(int i = 0;i < 256;i++)
                if(next[now][i] == -1)
                    next[now][i] = next[fail[now]][i];
                else
                {
                    fail[next[now][i]] = next[fail[now]][i];
                    Q.push(next[now][i]);
                }
        }
    }
    bool used[520];
    int query(unsigned char buf[],int len,int n)
    {
        memset(used,false,sizeof(used));
        int now = root;
        for(int i = 0;i < len;i++)
        {
            now = next[now][buf[i]];
            int temp = now;
            while( temp!=root )
            {
                if(end[temp] != -1)
                    used[end[temp]]=true;
                temp = fail[temp];
            }
        }
        int res = 0;
        for(int i = 0;i < n;i++)
            if(used[i])
                res++;
        return res;
    }
};

unsigned char buf[2050];
int tot;
char str[4000];
unsigned char s[4000];
unsigned char Get(char ch)
{
    if( ch>='A'&&ch<='Z' )return ch-'A';
    if( ch>='a'&&ch<='z' )return ch-'a'+26;
    if( ch>='0'&&ch<='9' )return ch-'0'+52;
    if( ch=='+' )return 62;
    else return 63;
}
void change(unsigned char str[],int len)
{
    int t=0;
    for(int i=0;i<len;i+=4)
    {
        buf[t++]=((str[i]<<2)|(str[i+1]>>4));
        if(i+2 < len)
            buf[t++]=( (str[i+1]<<4)|(str[i+2]>>2) );
        if(i+3 < len)
            buf[t++]= ( (str[i+2]<<6)|str[i+3] );
    }
    tot=t;
}
Trie ac;
int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    int n,m;
    while(scanf("%d",&n) == 1)
    {
        ac.init();
        for(int i = 0;i < n;i++)
        {
            scanf("%s",str);
            int len = strlen(str);
            while(str[len-1]=='=')len--;
            for(int j = 0;j < len;j++)
            {
                s[j] = Get(str[j]);
            }
            change(s,len);
            ac.insert(buf,tot,i);
        }
        ac.build();
        scanf("%d",&m);
        while(m--)
        {
            scanf("%s",str);
            int len=strlen(str);
            while(str[len-1]=='=')len--;
            for(int j = 0;j < len;j++)
                s[j] = Get(str[j]);
            change(s,len);
            printf("%d\n",ac.query(buf,tot,n));
        }
        printf("\n");
    }
    return 0;
}

View Code

5、POJ 2778 DNA Sequence

AC自动机+矩阵加速

这个时候AC自动机的一种状态转移图的思路就很透彻了。

AC自动机就是可以确定状态的转移。

//============================================================================
// Name        : POJ.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <queue>
using namespace std;

const int MOD=100000;
struct Matrix
{
    int mat[110][110],n;
    Matrix(){}
    Matrix(int _n)
    {
        n = _n;
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
                mat[i][j]=0;
    }
    Matrix operator *(const Matrix &b)const
    {
        Matrix ret=Matrix(n);
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
                for(int k=0;k<n;k++)
                {
                    int tmp=(long long)mat[i][k]*b.mat[k][j]%MOD;
                    ret.mat[i][j]=(ret.mat[i][j]+tmp)%MOD;
                }
        return ret;
    }
};
struct Trie
{
    int next[110][4],fail[110];
    bool end[110];
    int root,L;
    int newnode()
    {
        for(int i=0;i<4;i++)
            next[L][i]=-1;
        end[L++]=false;
        return L-1;
    }
    void init()
    {
        L=0;
        root=newnode();
    }
    int getch(char ch)
    {
        switch(ch)
        {
        case 'A':
            return 0;
            break;
        case 'C':
            return 1;
            break;
        case 'G':
            return 2;
            break;
        case 'T':
            return 3;
            break;
        }
    }
    void insert(char s[])
    {
        int len=strlen(s);
        int now=root;
        for(int i = 0;i < len;i++)
        {
            if(next[now][getch(s[i])] == -1)
                next[now][getch(s[i])] = newnode();
            now = next[now][getch(s[i])];
        }
        end[now]=true;
    }
    void build()
    {
        queue<int>Q;
        for(int i = 0;i < 4;i++)
            if(next[root][i] == -1)
                next[root][i] = root;
            else
            {
                fail[next[root][i]] = root;
                Q.push(next[root][i]);
            }
        while(!Q.empty())
        {
            int now = Q.front();
            Q.pop();
            if(end[fail[now]]==true)
                end[now]=true;
            for(int i = 0;i < 4;i++)
            {
                if(next[now][i] == -1)
                    next[now][i] = next[fail[now]][i];
                else
                {
                    fail[next[now][i]] = next[fail[now]][i];
                    Q.push(next[now][i]);
                }
            }
        }
    }
    Matrix getMatrix()
    {
        Matrix res = Matrix(L);
        for(int i=0;i<L;i++)
            for(int j=0;j<4;j++)
                if(end[next[i][j]]==false)
                    res.mat[i][next[i][j]]++;
        return res;
    }
};

Trie ac;
char buf[20];

Matrix pow_M(Matrix a,int n)
{
    Matrix ret = Matrix(a.n);
    for(int i = 0; i < ret.n; i++)
        ret.mat[i][i]=1;
    Matrix tmp=a;
    while(n)
    {
        if(n&1)ret=ret*tmp;
        tmp=tmp*tmp;
        n>>=1;
    }
    return ret;
}

int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m) != EOF)
    {
        ac.init();
        for(int i=0;i<n;i++)
        {
            scanf("%s",buf);
            ac.insert(buf);
        }
        ac.build();
        Matrix a=ac.getMatrix();
        a=pow_M(a,m);
        int ans=0;
        for(int i=0;i<a.n;i++)
        {
            ans=(ans+a.mat[0][i])%MOD;
        }
        printf("%d\n",ans);
    }
    return 0;
}

View Code

6、HDU 2243 考研路茫茫——单词情结

这题和上题有些类似。但是需要求和。

所以给矩阵增加一维，这样可以完美解决

题解here

//============================================================================
// Name        : HDU.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
using namespace std;
struct Matrix
{
    unsigned long long mat[40][40];
    int n;
    Matrix(){}
    Matrix(int _n)
    {
        n=_n;
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
                mat[i][j] = 0;
    }
    Matrix operator *(const Matrix &b)const
    {
        Matrix ret = Matrix(n);
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
                for(int k=0;k<n;k++)
                    ret.mat[i][j]+=mat[i][k]*b.mat[k][j];
        return ret;
    }
};
unsigned long long pow_m(unsigned long long a,int n)
{
    unsigned long long ret=1;
    unsigned long long tmp = a;
    while(n)
    {
        if(n&1)ret*=tmp;
        tmp*=tmp;
        n>>=1;
    }
    return ret;
}
Matrix pow_M(Matrix a,int n)
{
    Matrix ret = Matrix(a.n);
    for(int i=0;i<a.n;i++)
        ret.mat[i][i] = 1;
    Matrix tmp = a;
    while(n)
    {
        if(n&1)ret=ret*tmp;
        tmp=tmp*tmp;
        n>>=1;
    }
    return ret;
}
struct Trie
{
    int next[40][26],fail[40];
    bool end[40];
    int root,L;
    int newnode()
    {
        for(int i = 0;i < 26;i++)
            next[L][i] = -1;
        end[L++] = false;
        return L-1;
    }
    void init()
    {
        L = 0;
        root = newnode();
    }
    void insert(char buf[])
    {
        int len = strlen(buf);
        int now = root;
        for(int i = 0;i < len;i++)
        {
            if(next[now][buf[i]-'a'] == -1)
                next[now][buf[i]-'a'] = newnode();
            now = next[now][buf[i]-'a'];
        }
        end[now] = true;
    }
    void build()
    {
        queue<int>Q;
        fail[root]=root;
        for(int i = 0;i < 26;i++)
            if(next[root][i] == -1)
                next[root][i] = root;
            else
            {
                fail[next[root][i]] = root;
                Q.push(next[root][i]);
            }
        while(!Q.empty())
        {
            int now = Q.front();
            Q.pop();
            if(end[fail[now]])end[now]=true;
            for(int i = 0;i < 26;i++)
                if(next[now][i] == -1)
                    next[now][i] = next[fail[now]][i];
                else
                {
                    fail[next[now][i]] = next[fail[now]][i];
                    Q.push(next[now][i]);
                }
        }
    }
    Matrix getMatrix()
    {
        Matrix ret = Matrix(L+1);
        for(int i = 0;i < L;i++)
            for(int j = 0;j < 26;j++)
                if(end[next[i][j]]==false)
                    ret.mat[i][next[i][j]] ++;
        for(int i = 0;i < L+1;i++)
            ret.mat[i][L] = 1;
        return ret;
    }
    void debug()
    {
        for(int i = 0;i < L;i++)
        {
            printf("id = %3d,fail = %3d,end = %3d,chi = [",i,fail[i],end[i]);
            for(int j = 0;j < 26;j++)
                printf("%2d",next[i][j]);
            printf("]\n");
        }
    }
};
char buf[10];
Trie ac;
int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    int n,L;
    while(scanf("%d%d",&n,&L)==2)
    {
        ac.init();
        for(int i = 0;i < n;i++)
        {
            scanf("%s",buf);
            ac.insert(buf);
        }
        ac.build();
        Matrix a = ac.getMatrix();
        a = pow_M(a,L);
        unsigned long long res = 0;
        for(int i = 0;i < a.n;i++)
            res += a.mat[0][i];
        res--;

        /*
         * f[n]=1 + 26^1 + 26^2 +...26^n
         * f[n]=26*f[n-1]+1
         * {f[n] 1} = {f[n-1] 1}[26 0;1 1]
         * 数是f[L]-1;
         * 此题的L<2^31.矩阵的幂不能是L+1次，否则就超时了
         */
        a = Matrix(2);
        a.mat[0][0]=26;
        a.mat[1][0] = a.mat[1][1] = 1;
        a=pow_M(a,L);
        unsigned long long ans=a.mat[1][0]+a.mat[0][0];
        ans--;
        ans-=res;
        cout<<ans<<endl;
    }
    return 0;
}

View Code

7、POJ 1625 Censored!

AC自动机+DP+高精度

好题

题解here

//============================================================================
// Name        : POJ.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>
#include <queue>
#include <map>
using namespace std;
map<char,int>mp;
int N,M,P;
struct Matrix
{
    int mat[110][110];
    int n;
    Matrix(){}
    Matrix(int _n)
    {
        n=_n;
        for(int i = 0;i < n;i++)
            for(int j = 0;j < n;j++)
                mat[i][j] = 0;
    }
};
struct Trie
{
    int next[110][256],fail[110];
    bool end[110];
    int L,root;
    int newnode()
    {
        for(int i = 0;i < 256;i++)
            next[L][i] = -1;
        end[L++] = false;
        return L-1;
    }
    void init()
    {
        L = 0;
        root = newnode();
    }
    void insert(char buf[])
    {
        int len = strlen(buf);
        int now = root;
        for(int i = 0;i < len;i++)
        {
            if(next[now][mp[buf[i]]] == -1)
                next[now][mp[buf[i]]] = newnode();
            now = next[now][mp[buf[i]]];
        }
        end[now] = true;
    }
    void build()
    {
        queue<int>Q;
        fail[root] = root;
        for(int i = 0;i < 256;i++)
            if(next[root][i] == -1)
                next[root][i] = root;
            else
            {
                fail[next[root][i]] = root;
                Q.push(next[root][i]);
            }
        while(!Q.empty())
        {
            int now = Q.front();
            Q.pop();
            if(end[fail[now]]==true)end[now]=true;
            for(int i = 0;i < 256;i++)
                if(next[now][i] == -1)
                    next[now][i] = next[fail[now]][i];
                else
                {
                    fail[next[now][i]] = next[fail[now]][i];
                    Q.push(next[now][i]);
                }
        }
    }
    Matrix getMatrix()
    {
        Matrix res = Matrix(L);
        for(int i = 0;i < L;i++)
            for(int j = 0;j < N;j++)
                if(end[next[i][j]]==false)
                    res.mat[i][next[i][j]]++;
        return res;
    }
    void debug()
    {
        for(int i = 0;i < L;i++)
        {
            printf("id = %3d,fail = %3d,end = %3d,chi = [",i,fail[i],end[i]);
            for(int j = 0;j < 26;j++)
                printf("%2d",next[i][j]);
            printf("]\n");
        }
    }

};


/*
 * 高精度，支持乘法和加法
 */
struct BigInt
{
    const static int mod = 10000;
    const static int DLEN = 4;
    int a[600],len;
    BigInt()
    {
        memset(a,0,sizeof(a));
        len = 1;
    }
    BigInt(int v)
    {
        memset(a,0,sizeof(a));
        len = 0;
        do
        {
            a[len++] = v%mod;
            v /= mod;
        }while(v);
    }
    BigInt(const char s[])
    {
        memset(a,0,sizeof(a));
        int L = strlen(s);
        len = L/DLEN;
        if(L%DLEN)len++;
        int index = 0;
        for(int i = L-1;i >= 0;i -= DLEN)
        {
            int t = 0;
            int k = i - DLEN + 1;
            if(k < 0)k = 0;
            for(int j = k;j <= i;j++)
                t = t*10 + s[j] - '0';
            a[index++] = t;
        }
    }
    BigInt operator +(const BigInt &b)const
    {
        BigInt res;
        res.len = max(len,b.len);
        for(int i = 0;i <= res.len;i++)
            res.a[i] = 0;
        for(int i = 0;i < res.len;i++)
        {
            res.a[i] += ((i < len)?a[i]:0)+((i < b.len)?b.a[i]:0);
            res.a[i+1] += res.a[i]/mod;
            res.a[i] %= mod;
        }
        if(res.a[res.len] > 0)res.len++;
        return res;
    }
    BigInt operator *(const BigInt &b)const
    {
        BigInt res;
        for(int i = 0; i < len;i++)
        {
            int up = 0;
            for(int j = 0;j < b.len;j++)
            {
                int temp = a[i]*b.a[j] + res.a[i+j] + up;
                res.a[i+j] = temp%mod;
                up = temp/mod;
            }
            if(up != 0)
                res.a[i + b.len] = up;
        }
        res.len = len + b.len;
        while(res.a[res.len - 1] == 0 &&res.len > 1)res.len--;
        return res;
    }
    void output()
    {
        printf("%d",a[len-1]);
        for(int i = len-2;i >=0 ;i--)
            printf("%04d",a[i]);
        printf("\n");
    }
};
char buf[1010];
BigInt dp[2][110];
Trie ac;
int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);

    while(scanf("%d%d%d",&N,&M,&P)==3)
    {
        gets(buf);
        gets(buf);
        mp.clear();
        int len = strlen(buf);
        for(int i = 0;i < len;i++)
            mp[buf[i]]=i;
        ac.init();
        for(int i = 0;i < P;i++)
        {
            gets(buf);
            ac.insert(buf);
        }
        ac.build();
//        ac.debug();
        Matrix a= ac.getMatrix();

//        for(int i = 0;i <a.n;i++)
//        {
//            for(int j=0;j<a.n;j++)printf("%d ",a.mat[i][j]);
//            cout<<endl;
//        }

        int now = 0;
        dp[now][0] = 1;
        for(int i = 1;i < a.n;i++)
            dp[now][i] = 0;
        for(int i = 0;i < M;i++)
        {
            now^=1;
            for(int j = 0;j < a.n;j++)
                dp[now][j] = 0;
            for(int j = 0;j < a.n;j++)
                for(int k = 0;k < a.n;k++)
                    if(a.mat[j][k] > 0)
                        dp[now][k] = dp[now][k]+dp[now^1][j]*a.mat[j][k];
//            for(int j = 0;j < a.n;j++)
//                dp[now][j].output();
        }
        BigInt ans = 0;
        for(int i = 0;i < a.n;i++)
            ans = ans + dp[now][i];
        ans.output();
    }
    return 0;
}

View Code

8、HDU 2825 Wireless Password

AC自动机+状态压缩DP

相当于在AC自动机上产生状态转移，然后进行dp

//============================================================================
// Name        : HDU.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
using namespace std;
const int MOD=20090717;
int n,m,k;
int dp[30][110][1<<10];
int num[5000];

struct Trie
{
    int next[110][26],fail[110],end[110];
    int root,L;
    int newnode()
    {
        for(int i = 0;i < 26;i++)
            next[L][i] = -1;
        end[L++] = 0;
        return L-1;
    }
    void init()
    {
        L = 0;
        root = newnode();
    }
    void insert(char buf[],int id)
    {
        int len = strlen(buf);
        int now = root;
        for(int i = 0;i < len;i++)
        {
            if(next[now][buf[i]-'a']==-1)
                next[now][buf[i]-'a'] = newnode();
            now = next[now][buf[i]-'a'];
        }
        end[now] |= (1<<id);
    }
    void build()
    {
        queue<int>Q;
        fail[root] = root;
        for(int i = 0;i < 26;i++)
            if(next[root][i] == -1)
                next[root][i] = root;
            else
            {
                fail[next[root][i]] = root;
                Q.push(next[root][i]);
            }
        while(!Q.empty())
        {
            int now = Q.front();
            Q.pop();
            end[now] |= end[fail[now]];
            for(int i = 0;i < 26;i++)
                if(next[now][i] == -1)
                    next[now][i] = next[fail[now]][i];
                else
                {
                    fail[next[now][i]] = next[fail[now]][i];
                    Q.push(next[now][i]);
                }
        }
    }
    int solve()
    {
        //memset(dp,0,sizeof(dp));
        for(int i = 0;i <= n;i++)
            for(int j = 0;j < L;j++)
                for(int p = 0;p < (1<<m);p++)
                    dp[i][j][p]=0;
        dp[0][0][0] = 1;
        for(int i = 0;i < n;i++)
            for(int j = 0;j < L;j++)
                for(int p = 0;p< (1<<m);p++)
                    if(dp[i][j][p] > 0)
                    {
                        for(int x = 0;x < 26;x++)
                        {
                            int newi = i+1;
                            int newj = next[j][x];
                            int newp = (p|end[newj]);
                            dp[newi][newj][newp] += dp[i][j][p];
                            dp[newi][newj][newp]%=MOD;
                        }
                    }
        int ans = 0;
        for(int p = 0;p < (1<<m);p++)
        {
            if(num[p] < k)continue;
            for(int i = 0;i < L;i++)
            {
                ans = (ans + dp[n][i][p])%MOD;
            }
        }
        return ans;
    }
};
char buf[20];
Trie ac;
int main()
{
    for(int i=0;i<(1<<10);i++)
    {
        num[i] = 0;
        for(int j = 0;j < 10;j++)
            if(i & (1<<j))
                num[i]++;
    }
    while(scanf("%d%d%d",&n,&m,&k)==3)
    {
        if(n== 0 && m==0 &&k==0)break;
        ac.init();
        for(int i = 0;i < m;i++)
        {
            scanf("%s",buf);
            ac.insert(buf,i);
        }
        ac.build();
        printf("%d\n",ac.solve());
    }
    return 0;
}

View Code

9、HDU 2296 Ring

需要输出字典序最小的解

在DP的时候加一个字符数组来记录就行了

//============================================================================
// Name        : HDU.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
#include <queue>
using namespace std;

int a[110];
int dp[55][1110];
char str[55][1110][55];

bool cmp(char s1[],char s2[])
{
    int len1=strlen(s1);
    int len2=strlen(s2);
    if(len1 != len2)return len1 < len2;
    else return strcmp(s1,s2) < 0;
}

const int INF=0x3f3f3f3f;
struct Trie
{
    int next[1110][26],fail[1110],end[1110];
    int root,L;
    int newnode()
    {
        for(int i = 0;i < 26;i++)
            next[L][i] = -1;
        end[L++] = -1;
        return L-1;
    }
    void init()
    {
        L = 0;
        root = newnode();
    }
    void insert(char buf[],int id)
    {
        int len = strlen(buf);
        int now = root;
        for(int i = 0;i < len;i++)
        {
            if(next[now][buf[i]-'a'] == -1)
                next[now][buf[i]-'a'] = newnode();
            now = next[now][buf[i]-'a'];
        }
        end[now] = id;
    }
    void build()
    {
        queue<int>Q;
        fail[root] = root;
        for(int i = 0;i < 26;i++)
            if(next[root][i] == -1)
                next[root][i] = root;
            else
            {
                fail[next[root][i]] = root;
                Q.push(next[root][i]);
            }
        while(!Q.empty())
        {
            int now = Q.front();
            Q.pop();
            for(int i = 0;i < 26;i++)
                if(next[now][i] == -1)
                    next[now][i] = next[fail[now]][i];
                else
                {
                    fail[next[now][i]] = next[fail[now]][i];
                    Q.push(next[now][i]);
                }
        }
    }
    void solve(int n)
    {
        for(int i = 0;i <= n;i++)
            for(int j = 0;j < L;j++)
                dp[i][j] = -INF;
        dp[0][0] = 0;
        strcpy(str[0][0],"");
        char ans[55];
        strcpy(ans,"");
        int Max = 0;
        char tmp[55];
        for(int i = 0; i < n;i++)
            for(int j = 0;j < L;j++)
                if(dp[i][j]>=0)
                {
                    strcpy(tmp,str[i][j]);
                    int len = strlen(tmp);
                    for(int k = 0;k < 26;k++)
                    {
                        int nxt=next[j][k];
                        tmp[len] = 'a'+k;
                        tmp[len+1] = 0;
                        int tt = dp[i][j];
                        if(end[nxt] != -1)
                            tt+=a[end[nxt]];

                        if(dp[i+1][nxt]<tt || (dp[i+1][nxt]==tt && cmp(tmp,str[i+1][nxt])))
                        {
                            dp[i+1][nxt] = tt;
                            strcpy(str[i+1][nxt],tmp);
                            if(tt > Max ||(tt==Max && cmp(tmp,ans)))
                            {
                                Max = tt;
                                strcpy(ans,tmp);
                            }
                        }
                    }
                }
        printf("%s\n",ans);
    }
};
char buf[60];
Trie ac;
int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    int T;
    int n,m;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        ac.init();
        for(int i = 0;i < m;i++)
        {
            scanf("%s",buf);
            ac.insert(buf,i);
        }
        for(int i = 0;i < m;i++)
            scanf("%d",&a[i]);
        ac.build();
        ac.solve(n);
    }
    return 0;
}

View Code

10、HDU 2457 DNA repair

很简单的AC自动机+DP了

//============================================================================
// Name        : HDU.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
#include <queue>
using namespace std;
const int INF = 0x3f3f3f3f;
struct Trie
{
    int next[1010][4],fail[1010];
    bool end[1010];
    int root,L;
    int newnode()
    {
        for(int i = 0;i < 4;i++)
            next[L][i] = -1;
        end[L++] = false;
        return L-1;
    }
    void init()
    {
        L = 0;
        root = newnode();
    }
    int getch(char ch)
    {
        if(ch == 'A')return 0;
        else if(ch == 'C')return 1;
        else if(ch == 'G')return 2;
        else if(ch == 'T')return 3;
    }
    void insert(char buf[])
    {
        int len = strlen(buf);
        int now = root;
        for(int i = 0;i < len;i++)
        {
            if(next[now][getch(buf[i])] == -1)
                next[now][getch(buf[i])] = newnode();
            now = next[now][getch(buf[i])];
        }
        end[now] = true;
    }
    void build()
    {
        queue<int>Q;
        fail[root] = root;
        for(int i = 0;i < 4;i++)
            if(next[root][i] == -1)
                next[root][i] = root;
            else
            {
                fail[next[root][i]] = root;
                Q.push(next[root][i]);
            }
        while(!Q.empty())
        {
            int now = Q.front();
            Q.pop();
            if(end[fail[now]])end[now] = true;//这里不要忘记
            for(int i = 0;i < 4;i++)
                if(next[now][i] == -1)
                    next[now][i] = next[fail[now]][i];
                else
                {
                    fail[next[now][i]] = next[fail[now]][i];
                    Q.push(next[now][i]);
                }
        }
    }
    int dp[1010][1010];
    int solve(char buf[])
    {
        int len = strlen(buf);
        for(int i = 0;i <= len;i++)
            for(int j = 0;j < L;j++)
                dp[i][j] = INF;
        dp[0][root] = 0;
        for(int i = 0;i < len;i++)
            for(int j = 0;j < L;j++)
                if(dp[i][j] < INF)
                {
                    for(int k = 0;k < 4;k++)
                    {
                        int news = next[j][k];
                        if(end[news])continue;
                        int tmp;
                        if( k == getch(buf[i]))tmp = dp[i][j];
                        else tmp = dp[i][j] + 1;
                        dp[i+1][news] = min(dp[i+1][news],tmp);
                    }
                }
        int ans = INF;
        for(int j = 0;j < L;j++)
            ans = min(ans,dp[len][j]);
        if(ans == INF)ans = -1;
        return ans;
    }

};
char buf[1010];
Trie ac;
int main()
{
    int n;
    int iCase = 0;
    while ( scanf("%d",&n) == 1 && n)
    {
        iCase++;
        ac.init();
        while(n--)
        {
            scanf("%s",buf);
            ac.insert(buf);
        }
        ac.build();
        scanf("%s",buf);
        printf("Case %d: %d\n",iCase,ac.solve(buf));
    }
    return 0;
}

View Code

11、ZOJ 3228 Searching the String

这题需要查询两种，一种是可重叠，一种是不可重叠的。

找模式串在目标串中的出现次数。

加一个数组记录上一次出现的位置，然后就可以求出不可重叠的了

//============================================================================
// Name        : ZOJ.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
using namespace std;

struct Trie
{
    int next[600010][26],fail[600010],deep[600010];
    int root,L;
    int newnode()
    {
        for(int i = 0;i < 26;i++)
            next[L][i] = -1;
        L++;
        return L-1;
    }
    void init()
    {
        L = 0;
        root = newnode();
        deep[root] = 0;
    }
    int insert(char buf[])
    {
        int len = strlen(buf);
        int now = root;
        for(int i = 0;i < len;i++)
        {
            if(next[now][buf[i]-'a'] == -1)
            {
                next[now][buf[i]-'a'] = newnode();
                deep[ next[now][buf[i]-'a'] ] = i+1;
            }
            now = next[now][buf[i]-'a'];
        }
        return now;
    }
    void build()
    {
        queue<int>Q;
        fail[root] = root;
        for(int i = 0;i < 26;i++)
            if(next[root][i] == -1)
                next[root][i] = root;
            else
            {
                fail[next[root][i]] = root;
                Q.push(next[root][i]);
            }
        while(!Q.empty())
        {
            int now = Q.front();
            Q.pop();
            for(int i = 0;i < 26;i++)
                if(next[now][i] == -1)
                    next[now][i] = next[fail[now]][i];
                else
                {
                    fail[next[now][i]] = next[fail[now]][i];
                    Q.push(next[now][i]);
                }
        }
    }
    int cnt[600010][2];
    int last[600010];
    void query(char buf[])
    {
        int len = strlen(buf);
        int now = root;
        memset(cnt,0,sizeof(cnt));
        memset(last,-1,sizeof(last));
        for(int i = 0;i < len;i++)
        {
            now = next[now][buf[i]-'a'];
            int temp = now;
            while(temp != root)
            {
                cnt[temp][0]++;
                if(i - last[temp] >= deep[temp])
                {
                    last[temp] = i;
                    cnt[temp][1]++;
                }
                temp = fail[temp];
            }
        }
    }
};
Trie ac;
char str[100010];
char buf[20];
int typ[100010],pos[100010];
int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    int n;
    int iCase = 0;
    while(scanf("%s",str)==1)
    {
        iCase++;
        printf("Case %d\n",iCase);
        scanf("%d",&n);
        ac.init();
        for(int i = 0;i < n;i++)
        {
            scanf("%d%s",&typ[i],buf);
            pos[i]=ac.insert(buf);
        }
        ac.build();
        ac.query(str);
        for(int i = 0;i < n;i++)
            printf("%d\n",ac.cnt[pos[i]][typ[i]]);
        printf("\n");
    }
    return 0;
}

View Code

12、HDU 3341 Lost's revenge

这题主要是状态的表示，就是记录ACGT出现的次数。

然后这个ACGT次数表示的时候，状态要转化。

题解here

//============================================================================
// Name        : HDU.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
#include <queue>
using namespace std;
const int INF = 0x3f3f3f3f;
struct Trie
{
    int next[510][4],fail[510];
    int end[510];
    int root,L;
    int newnode()
    {
        for(int i = 0;i < 4;i++)
            next[L][i] = -1;
        end[L++] = 0;
        return L-1;
    }
    void init()
    {
        L = 0;
        root = newnode();
    }
    int getch(char ch)
    {
        if(ch == 'A')return 0;
        else if(ch == 'C')return 1;
        else if(ch == 'G')return 2;
        else return 3;
    }
    void insert(char buf[])
    {
        int len = strlen(buf);
        int now = root;
        for(int i = 0;i < len;i++)
        {
            if(next[now][getch(buf[i])] == -1)
                next[now][getch(buf[i])] = newnode();
            now = next[now][getch(buf[i])];
        }
        end[now] ++;
    }
    void build()
    {
        queue<int>Q;
        fail[root] = root;
        for(int i = 0;i < 4;i++)
            if(next[root][i] == -1)
                next[root][i] = root;
            else
            {
                fail[next[root][i]] = root;
                Q.push(next[root][i]);
            }
        while(!Q.empty())
        {
            int now = Q.front();
            Q.pop();
            end[now] += end[fail[now]];/********/
            for(int i = 0;i < 4;i++)
                if(next[now][i] == -1)
                    next[now][i] = next[fail[now]][i];
                else
                {
                    fail[next[now][i]] = next[fail[now]][i];
                    Q.push(next[now][i]);
                }
        }
    }
    int dp[510][11*11*11*11+5];
    int bit[4];
    int num[4];
    int solve(char buf[])
    {
        int len = strlen(buf);
        memset(num,0,sizeof(num));
        for(int i = 0;i < len;i++)
            num[getch(buf[i])]++;
        bit[0] = (num[1]+1)*(num[2]+1)*(num[3]+1);
        bit[1] = (num[2]+1)*(num[3]+1);
        bit[2] = (num[3]+1);
        bit[3] = 1;
        memset(dp,-1,sizeof(dp));
        dp[root][0] = 0;
        for(int A = 0;A <= num[0];A++)
            for(int B = 0;B <= num[1];B++)
                for(int C = 0;C <= num[2];C++)
                    for(int D = 0;D <= num[3];D++)
                    {
                        int s = A*bit[0] + B*bit[1] + C*bit[2] + D*bit[3];
                        for(int i = 0;i < L;i++)
                            if(dp[i][s] >= 0)
                            {
                                for(int k = 0;k < 4;k++)
                                {
                                    if(k == 0 && A == num[0])continue;
                                    if(k == 1 && B == num[1])continue;
                                    if(k == 2 && C == num[2])continue;
                                    if(k == 3 && D == num[3])continue;
                                    dp[next[i][k]][s+bit[k]] = max(dp[next[i][k]][s+bit[k]],dp[i][s]+end[next[i][k]]);
                                }
                            }
                    }
        int ans = 0;
        int status = num[0]*bit[0] + num[1]*bit[1] + num[2]*bit[2] + num[3]*bit[3];
        for(int i = 0;i < L;i++)
            ans = max(ans,dp[i][status]);
        return ans;
    }
};
char buf[50];
Trie ac;
int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    int n;
    int iCase = 0;
    while( scanf("%d",&n) == 1 && n)
    {
        iCase++;
        ac.init();
        for(int i = 0;i < n;i++)
        {
            scanf("%s",buf);
            ac.insert(buf);
        }
        ac.build();
        scanf("%s",buf);
        printf("Case %d: %d\n",iCase,ac.solve(buf));
    }
    return 0;
}

View Code

13、HDU 3247 Resource Archiver

使用最短路预处理出状态的转移。这样可以优化很多

//============================================================================
// Name        : HDU.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
using namespace std;

const int INF = 0x3f3f3f3f;

struct Trie
{
    int next[60010][2],fail[60010],end[60010];
    int root,L;
    int newnode()
    {
        for(int i = 0;i < 2;i++)
            next[L][i] = -1;
        end[L++] = 0;
        return L-1;
    }
    void init()
    {
        L = 0;
        root = newnode();
    }
    void insert(char buf[],int id)
    {
        int len = strlen(buf);
        int now = root;
        for(int i = 0;i < len ;i++)
        {
            if(next[now][buf[i]-'0'] == -1)
                next[now][buf[i]-'0'] = newnode();
            now = next[now][buf[i]-'0'];
        }
        end[now] = id;
    }
    void build()
    {
        queue<int>Q;
        fail[root] = root;
        for(int i = 0;i < 2;i++)
            if(next[root][i] == -1)
                next[root][i] = root;
            else
            {
                fail[next[root][i]] = root;
                Q.push(next[root][i]);
            }
        while( !Q.empty() )
        {
            int now = Q.front();
            Q.pop();
            if(end[fail[now]] == -1)end[now] = -1;
            else end[now] |= end[fail[now]];
            for(int i = 0;i < 2;i++)
                if(next[now][i] == -1)
                    next[now][i] = next[fail[now]][i];
                else
                {
                    fail[next[now][i]] = next[fail[now]][i];
                    Q.push(next[now][i]);
                }
        }
    }
    int g[11][11];
    int dp[1025][11];
    int cnt;
    int pos[11];
    int dis[60010];


    void bfs(int k)
    {
        queue<int>q;
        memset(dis,-1,sizeof(dis));
        dis[pos[k]] = 0;
        q.push(pos[k]);
        while(!q.empty())
        {
            int now = q.front();
            q.pop();
            for(int i = 0; i< 2;i++)
            {
                int tmp = next[now][i];
                if(dis[tmp]<0 && end[tmp] >= 0)
                {
                    dis[tmp] = dis[now] + 1;
                    q.push(tmp);
                }
            }
        }
        for(int i = 0;i < cnt;i++)
            g[k][i] = dis[pos[i]];
    }


    int solve(int n)
    {

        pos[0] = 0;
        cnt = 1;
        for(int i = 0;i < L;i++)
            if(end[i] > 0)
                pos[cnt++] = i;
        for(int i = 0; i < cnt;i++)
            bfs(i);

        for(int i = 0;i < (1<<n);i++)
            for(int j = 0;j < cnt;j++)
                dp[i][j] = INF;
        dp[0][0] = 0;
        for(int i = 0;i <(1<<n);i++)
            for(int j = 0;j < cnt;j++)
                if(dp[i][j]<INF)
                {
                    for(int k = 0;k < cnt;k++)
                    {
                        if(g[j][k] < 0)continue;
                        if( j == k)continue;
                        dp[i|end[pos[k]]][k] = min(dp[i|end[pos[k]]][k],dp[i][j]+g[j][k]);
                    }
                }
        int ans = INF;
        for(int j = 0;j < cnt;j++)
            ans = min(ans,dp[(1<<n)-1][j]);
        return ans;
    }
};
Trie ac;
char buf[1010];

int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    int n,m;
    while(scanf("%d%d",&n,&m) == 2)
    {
        if(n == 0 && m == 0)break;
        ac.init();
        for(int i = 0;i < n;i++)
        {
            scanf("%s",buf);
            ac.insert(buf,1<<i);
        }
        for(int i = 0;i < m;i++)
        {
            scanf("%s",buf);
            ac.insert(buf,-1);
        }
        ac.build();
        printf("%d\n",ac.solve(n));
    }
    return 0;
}

View Code

14、ZOJ 3494 BCD Code

这道题很神，数位DP和AC自动机结合，太强大了。

题解here

//============================================================================
// Name        : ZOJ.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
using namespace std;

struct Trie
{
    int next[2010][2],fail[2010];
    bool end[2010];
    int root,L;
    int newnode()
    {
        for(int i = 0;i < 2;i++)
            next[L][i] = -1;
        end[L++] = false;
        return L-1;
    }
    void init()
    {
        L = 0;
        root = newnode();
    }
    void insert(char buf[])
    {
        int len = strlen(buf);
        int now = root;
        for(int i = 0;i < len ;i++)
        {
            if(next[now][buf[i]-'0'] == -1)
                next[now][buf[i]-'0'] = newnode();
            now = next[now][buf[i]-'0'];
        }
        end[now] = true;
    }
    void build()
    {
        queue<int>Q;
        fail[root] = root;
        for(int i = 0;i < 2;i++)
            if(next[root][i] == -1)
                next[root][i] = root;
            else
            {
                fail[next[root][i]] = root;
                Q.push(next[root][i]);
            }
        while(!Q.empty())
        {
            int now = Q.front();
            Q.pop();
            if(end[fail[now]])end[now] = true;
            for(int i = 0;i < 2;i++)
                if(next[now][i] == -1)
                    next[now][i] = next[fail[now]][i];
                else
                {
                    fail[next[now][i]] = next[fail[now]][i];
                    Q.push(next[now][i]);
                }
        }
    }
};
Trie ac;

int bcd[2010][10];
int change(int pre,int num)
{
    if(ac.end[pre])return -1;
    int cur = pre;
    for(int i = 3;i >= 0;i--)
    {
        if(ac.end[ac.next[cur][(num>>i)&1]])return -1;
        cur = ac.next[cur][(num>>i)&1];
    }
    return cur;
}
void pre_init()
{
    for(int i = 0;i <ac.L;i++)
        for(int j = 0;j <10;j++)
            bcd[i][j] = change(i,j);
}
const int MOD = 1000000009;
long long dp[210][2010];
int bit[210];

long long dfs(int pos,int s,bool flag,bool z)
{
    if(pos == -1)return 1;
    if(!flag && dp[pos][s]!=-1)return dp[pos][s];
    long long ans = 0;
    if(z)
    {
        ans += dfs(pos-1,s,flag && bit[pos]==0,true);
        ans %= MOD;
    }
    else
    {
        if(bcd[s][0]!=-1)ans += dfs(pos-1,bcd[s][0],flag && bit[pos]==0,false);
        ans %= MOD;
    }
    int end = flag?bit[pos]:9;
    for(int i = 1;i<=end;i++)
    {
        if(bcd[s][i]!=-1)
        {
            ans += dfs(pos-1,bcd[s][i],flag&&i==end,false);
            ans %=MOD;
        }
    }
    if(!flag && !z)dp[pos][s] = ans;
    return ans;
}

long long calc(char s[])
{
    int len = strlen(s);
    for(int i = 0;i < len;i++)
        bit[i] = s[len-1-i]-'0';
    return dfs(len-1,0,1,1);
}
char str[210];
int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    int T;
    scanf("%d",&T);
    int n;
    while(T--)
    {
        ac.init();
        scanf("%d",&n);
        for(int i = 0;i < n;i++)
        {
            scanf("%s",str);
            ac.insert(str);
        }
        ac.build();
        pre_init();
        memset(dp,-1,sizeof(dp));
        int ans = 0;
        scanf("%s",str);
        int len = strlen(str);
        for(int i = len -1;i >=0;i--)
        {
            if(str[i]>'0')
            {
                str[i]--;
                break;
            }
            else str[i] = '9';
        }
        ans -= calc(str);
        ans %=MOD;
        scanf("%s",str);
        ans += calc(str);
        ans %=MOD;
        if(ans < 0)ans += MOD;
        printf("%d\n",ans);
    }
    return 0;
}

View Code

先简单总结到这吧！抱歉，时间原因，写得很简单，以后有机会好好补充完全吧！

posted on 2013-06-30 23:16 kuangbin 阅读(30680) 评论(5) 编辑收藏举报

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