【POJ2778】AC自动机+矩阵乘法

DNA Sequence

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 14758	Accepted: 5716

Description

It's well known that DNA Sequence is a sequence only contains A, C, T and G, and it's very useful to analyze a segment of DNA Sequence，For example, if a animal's DNA sequence contains segment ATC then it may mean that the animal may have a genetic disease. Until now scientists have found several those segments, the problem is how many kinds of DNA sequences of a species don't contain those segments. 

Suppose that DNA sequences of a species is a sequence that consist of A, C, T and G，and the length of sequences is a given integer n. 

Input

First line contains two integer m (0 <= m <= 10), n (1 <= n <=2000000000). Here, m is the number of genetic disease segment, and n is the length of sequences. 

Next m lines each line contain a DNA genetic disease segment, and length of these segments is not larger than 10. 

Output

An integer, the number of DNA sequences, mod 100000.

Sample Input

4 3
AT
AC
AG
AA

Sample Output

36

【分析】
　　之前做过的题。  建立一个AC自动机，然后把每个字符串的最后一位所在的节点标黑。（就是把mark变为1） 然后用AC自动机建立矩阵array[i][j]表示从i走到j的方案数，然后利用矩阵乘法的性质，用快速幂求出矩阵的M次幂即可。


代码依然丑：

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
#define Maxn 1010
#define LL long long
#define Mod 100000

struct node
{
    int x,fail;
    int son[6];
    bool mark;
}t[Maxn];int tot=0;

struct Matrix
{
    LL a[110][110];
}ay,ut;

void upd(int x)
{
    t[x].mark=0;
    memset(t[x].son,0,sizeof(t[x].son));
}

Matrix h;
void clear(int id)
{
    for(int i=0;i<=tot;i++)
     for(int j=0;j<=tot;j++)
      if(id==1) ay.a[i][j]=0;
      else h.a[i][j]=0;
}

int m,n;
char s[Maxn];

void read_trie()
{
    scanf("%s",s+1);
    int len=strlen(s+1);
    int now=0;
    for(int i=1;i<=len;i++)
    {
        int ind;
        if(s[i]=='A') ind=1;
        else if(s[i]=='T') ind=2;
        else if(s[i]=='C') ind=3;
        else ind=4;
        if(!t[now].son[ind])
        {
            t[now].son[ind]=++tot,upd(tot);
        }
        now=t[now].son[ind];
        if(i==len) t[now].mark=1;
    }
}

queue<int > q;
void build_AC()
{
    while(!q.empty()) q.pop();
    q.push(0);
    while(!q.empty())
    {
        int now=q.front();q.pop();
        int f=t[now].fail;
        for(int i=1;i<=4;i++)
        {
            if(t[now].son[i])
            {
                q.push(t[now].son[i]);
                t[t[now].son[i]].fail=now?t[f].son[i]:0;
                if(t[t[t[now].son[i]].fail].mark) t[t[now].son[i]].mark=1;
            }
            else t[now].son[i]=t[f].son[i];
            if(t[t[now].son[i]].mark!=1) ay.a[now][t[now].son[i]]++;
        }
    }
}

Matrix mul(Matrix a,Matrix b)
{
    clear(2);
    for(int i=0;i<=tot;i++)
     for(int j=0;j<=tot;j++)
      for(int k=0;k<=tot;k++)
      {
        h.a[i][j]=(h.a[i][j]+a.a[i][k]*b.a[k][j])%Mod;
          
      }
    return h;
}

void qpow(int k)
{
    while(k)
    {
        if(k&1) ut=mul(ut,ay);
        ay=mul(ay,ay);
        k>>=1;
    }
}

void init()
{
    scanf("%d%d",&m,&n);
    tot=0;upd(0);
    for(int i=1;i<=m;i++)
    {
        read_trie();
    }
    clear(1);
    build_AC();
    for(int i=0;i<=tot;i++)
     for(int j=0;j<=tot;j++)
        ut.a[i][j]=i==j?1:0;
     
    qpow(n);
    LL ans=0;
    for(int i=0;i<=tot;i++) ans=(ans+ut.a[0][i])%Mod;
    printf("%I64d\n",ans);
}

int main()
{
    init();
    return 0;
}

 

 以前的code：

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<queue>
using namespace std;

const int Maxn=150;
const int Mod=(int)1e5;
char s[Maxn];
int q[Maxn];
long long ay[Maxn][Maxn],sum[Maxn][Maxn];
int n,m,tot,len;

struct node
{
    int son[6],cnt,fail,mark;
}t[Maxn];

void floy()
{
    tot=0;
    for(int i=1;i<=Maxn;i++)
    {
        t[i].cnt=0;t[i].mark=0;
        for(int j=1;j<=5;j++) t[i].son[j]=0;
    }
    memset(q,0,sizeof(q));
}

void read_trie()
{
    tot=0;
    int i,j;
    scanf("%d%d",&m,&n);
    for(i=1;i<=m;i++)
    {
        int x,ind;
        scanf("%s",s+1);
        len=strlen(s+1);
        x=0;
        for(j=1;j<=len;j++)
        {
            if(s[j]=='A') ind=1;
            else if(s[j]=='C') ind=2;
            else if(s[j]=='T') ind=3;
            else if(s[j]=='G') ind=4;
            if(!t[x].son[ind]) t[x].son[ind]=++tot;
            x=t[x].son[ind];
            t[x].cnt++;
            if(j==len) t[x].mark=1;
        }
    }
}

void build_AC()
{
    int i,j,x,y;
    q[++q[0]]=0;
    //for(i=1;i<=4;i++)
      //if(t[0].son[i]) q[++q[0]]=t[0].son[i];
    for(i=1;i<=q[0];i++)
    {
        x=q[i];
        y=t[x].fail;
        for(j=1;j<=4;j++)
        {
            if(t[x].son[j])
            {
                //t[t[x].son[j]].fail=t[y].son[j];
                q[++q[0]]=t[x].son[j];
                 t[t[x].son[j]].fail=x?t[y].son[j]:x;  
                if(t[t[t[x].son[j]].fail].mark) t[t[x].son[j]].mark=1;
            }
            else t[x].son[j]=t[y].son[j];
            if(!t[t[x].son[j]].mark) ay[x][t[x].son[j]]++;
        }
    }
}

void MatrixMult(long long a[Maxn][Maxn],long long b[Maxn][Maxn])
{
    int i,j,k;
    long long c[Maxn][Maxn];
    memset(c,0,sizeof(c));
    for(i=0;i<=tot;i++)
     for(j=0;j<=tot;j++)
      for(k=0;k<=tot;k++)
       c[i][j]+=a[i][k]*b[k][j];
    for(i=0;i<=tot;i++)
     for(j=0;j<=tot;j++)
      a[i][j]=c[i][j]%Mod;
}

long long MatrixPow(int k)
{
    int i,j;
    for(i=0;i<=tot;i++)
     for(j=0;j<=tot;j++)
       sum[i][j]=(i==j);
    while(k)
    {
        if(k&1) MatrixMult(sum,ay);
        MatrixMult(ay,ay);
        k>>=1;
    }
    long long ans=0;
    for(i=0;i<=tot;i++) ans+=sum[0][i];
    return ans%Mod;
}

int main()
{
    floy();
    read_trie();
    build_AC();
    printf("%I64d\n",MatrixPow(n));
    return 0;
}

 

 

2016-06-23 16:22:32