poj1625 ac自动机

Censored!

Time Limit: 5000MS		Memory Limit: 10000K
Total Submissions: 8102		Accepted: 2191

Description

The alphabet of Freeland consists of exactly N letters. Each sentence of Freeland language (also known as Freish) consists of exactly M letters without word breaks. So, there exist exactly N^M different Freish sentences. 

But after recent election of Mr. Grass Jr. as Freeland president some words offending him were declared unprintable and all sentences containing at least one of them were forbidden. The sentence S contains a word W if W is a substring of S i.e. exists such k >= 1 that S[k] = W[1], S[k+1] = W[2], ...,S[k+len(W)-1] = W[len(W)], where k+len(W)-1 <= M and len(W) denotes length of W. Everyone who uses a forbidden sentence is to be put to jail for 10 years. 

Find out how many different sentences can be used now by freelanders without risk to be put to jail for using it. 

Input

The first line of the input file contains three integer numbers: N -- the number of letters in Freish alphabet, M -- the length of all Freish sentences and P -- the number of forbidden words (1 <= N <= 50, 1 <= M <= 50, 0 <= P <= 10). 

The second line contains exactly N different characters -- the letters of the Freish alphabet (all with ASCII code greater than 32). 

The following P lines contain forbidden words, each not longer than min(M, 10) characters, all containing only letters of Freish alphabet. 

Output

Output the only integer number -- the number of different sentences freelanders can safely use.

Sample Input

2 3 1
ab
bb

Sample Output

5

 

 

在AC自动机上dp,dp[i][j]表示走了i位，在ac自动机上第j个位置的方案数。ans比较大，要用高精度。

据说有些鬼畜的数据。所以不要用s[i]-32，用hash比较好

 

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<cstring>

using namespace std;

struct node{
    node *fail;
    node *son[100];
    int num;
    bool ar;
    node()
    {
        fail=NULL;
        ar=true;
        num=0;
        for(int i=0;i<100;i++)son[i]=NULL;
    }
} *que[1011];
node *root;
char s[1011];
int len,n,m,t,i,tot;
node *ac[1011];
struct ans{
    int n;
    int a[101];
}dp[61][601];
ans xzq;
int hash[601];

ans plus(ans a,ans b)
{
    ans c;
    int i;
    for(i=0;i<=100;i++)c.a[i]=0;
    if(a.n>b.n)c.n=a.n;
    else c.n=b.n;
    for(i=1;i<=c.n;i++){
        c.a[i]=c.a[i]+a.a[i]+b.a[i];
        if(c.a[i]>=10){
            c.a[i+1]=c.a[i+1]+c.a[i]/10;
            c.a[i]=c.a[i]%10;
        }
    }
    c.n++;
    if(c.a[c.n]==0)c.n--;
    return c;
}

void Read()
{
    len=0;
    char c;
    while(c=getchar(),c=='\n');
    s[++len]=c;
    while(c=getchar(),c!='\n')s[++len]=c;
}

void add()
{
    int i,ws;
    node *p;
    p=root;
    for(i=1;i<=len;i++){
        ws=hash[s[i]];
        if(p->son[ws]==NULL){
            p->son[ws]=new node();
            p->son[ws]->num=++tot;
            ac[tot]=p->son[ws];
        }
        p=p->son[ws];
    }
    p->ar=false;
}

void Bfs()
{
    int l,r,i,j,k;
    node *p;
    node *q;
    que[l=r=1]=root;
    while(l<=r){
        p=que[l];
        for(i=1;i<=n;i++)if(p->son[i]!=NULL){
            r++;
            que[r]=p->son[i];
            if(p==root)p->son[i]->fail=root;
            else{
                q=p->fail;
                while(q!=NULL){
                    if(q->son[i]!=NULL){
                        p->son[i]->fail=q->son[i];
                        break;
                    }
                    q=q->fail;
                }
                if(p->son[i]->fail==NULL)p->son[i]->fail=root;
            }
        }
        l++;
    }
    for(i=2;i<=r;i++)que[i]->ar&=que[i]->fail->ar;
}

void Main()
{
    int i,j,k;
    node *p;
    node *q;
    dp[0][1].n=1;
    dp[0][1].a[1]=1;
    for(i=0;i<m;i++){
        for(j=1;j<=tot;j++){
            for(k=1;k<=n;k++){
                p=ac[j];
                if(p->son[k]!=NULL)p=p->son[k];
                else{
                    q=p->fail;
                    while(q!=NULL){
                        if(q->son[k]!=NULL){
                            p=q->son[k];
                            break;
                        }
                        q=q->fail;
                    }
                    if(q==NULL)p=root;
                }
                if(p->ar==true)dp[i+1][p->num]=plus(dp[i+1][p->num],dp[i][j]);
            }
        }
    }
    for(i=1;i<=tot;i++)xzq=plus(xzq,dp[m][i]);
}

int main()
{
    tot=0;
    root=new node();
    root->num=++tot;
    ac[tot]=root;
    scanf("%d%d%d",&n,&m,&t);
    Read();
    for(i=1;i<=len;i++)hash[s[i]]=i;
    for(i=1;i<=t;i++){
        Read();
        add();
    }
    Bfs();
    Main();
    if(xzq.n==0)printf("0\n");
    else{
        for(i=xzq.n;i>=1;i--)printf("%d",xzq.a[i]);
        printf("\n");
    }
}