AC自动机构造字符串（好）poj2778

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DNA Sequence

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11630 Accepted: 4439

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 3ATACAGAA

Sample Output

36

题意：问长度为n的，并且不包含给出的任意一个字符串的串有多少个

思路：首先建立ac自动机，然后构造转移矩阵，下面的博客讲得很清楚，加一点自己的理解在代码里。

http://hi.baidu.com/ccsu_010/item/7847a3c17f6fe2bc0d0a7b89

#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;typedef long long LL;const int maxn=15;const int maxm=15*10;const int SIGMA_SIZE=4;const int MOD=100000;char word[20];int n,m,sz;int cnt[110][110];struct Matrix{    int mat[110][110];    Matrix(){memset(mat,0,sizeof(mat));}    Matrix operator*(Matrix a)    {        Matrix res;        for(int k=0;k<sz;k++)        for(int i=0;i<sz;i++)        {            if(mat[i][k]==0)continue;            for(int j=0;j<sz;j++)                if(a.mat[k][j])                    res.mat[i][j]=(res.mat[i][j]+(LL)mat[i][k]*a.mat[k][j]%MOD)%MOD;        }        return res;    }};struct AC{    int ch[maxm][4],val[maxm];    int fail[maxm],last[maxm];    void clear(){memset(ch[0],0,sizeof(ch[0]));sz=1;val[0]=0;}    int idx(char x)    {        if(x=='A')return 0;        else if(x=='C')return 1;        else if(x=='T')return 2;        return 3;    }    void insert(char *s)    {        int u=0;        int n=strlen(s);        for(int i=0;i<n;i++)        {            int c=idx(s[i]);            if(!ch[u][c])            {                memset(ch[sz],0,sizeof(ch[sz]));                val[sz]=0;                ch[u][c]=sz++;            }            u=ch[u][c];        }        val[u]=1;    }    void getfail()    {        queue<int> q;        int u=0;        fail[0]=0;        for(int i=0;i<SIGMA_SIZE;i++)        {            u=ch[0][i];            if(u){fail[u]=last[u]=0;q.push(u);}        }        while(!q.empty())        {            int r=q.front();q.pop();            if(val[fail[r]])val[r]=1;//失配指针相连的，若果fail[u]是单词节点，那么相应的当前节点也是单词节点，因为fail[u]是这个的前缀            for(int c=0;c<SIGMA_SIZE;c++)            {                u=ch[r][c];                if(!u){ch[r][c]=ch[fail[r]][c];continue;}                q.push(u);                int v=fail[r];                while(v&&!ch[v][c])v=fail[v];                fail[u]=ch[v][c];                last[u]=val[fail[u]]?fail[u]:last[fail[u]];            }        }    }    Matrix getMatrix()//根据是否是单词节点构造矩阵，进行转移    {        Matrix res;        for(int i=0;i<sz;i++)            for(int j=0;j<4;j++)                if(!val[ch[i][j]])                    res.mat[i][ch[i][j]]++;        return res;    }}ac;Matrix pow_mul(Matrix A,int x){    Matrix res;    for(int i=0;i<=sz;i++)res.mat[i][i]=1;    while(x)    {        if(x&1)res=res*A;        A=A*A;        x>>=1;    }    return res;}int main(){    while(scanf("%d%d",&n,&m)!=EOF)    {        ac.clear();        for(int i=1;i<=n;i++)        {            scanf("%s",word);            ac.insert(word);        }        ac.getfail();        Matrix A=ac.getMatrix();        A=pow_mul(A,m);        int ans=0;        for(int i=0;i<sz;i++)            ans=(ans+A.mat[0][i])%MOD;        printf("%d\n",ans);    }    return 0;}