DNA Sequence

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 11721	Accepted: 4471

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

  給M個串，問有多少個長度為N的的串不包含給出的M個串。

  首先把M個串建立AC自動機，然後構造一個大小為ac.size*ac.size的矩陣，mat[i][j]代表節點i走到節點j走一步的合法路徑數，i和j應該是合法節點。這樣形成了一個一步矩陣，N步的話把這個矩陣N次方，答案是矩陣第一行的和。

#include<iostream>
#include<queue>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<algorithm>
#define INF 0x3f3f3f3f
#define eps 1e-9
#define MAXN 60010
#define MAXM 2000010
#define MAXNODE 105
#define MOD 100000
#define SIGMA_SIZE 4
typedef long long LL;
using namespace std;
int T,M,N;
char str[15];
struct AC{
    int ch[MAXNODE][SIGMA_SIZE],f[MAXNODE],val[MAXNODE],sz;
    void init(){
        memset(ch[0],0,sizeof(ch[0]));
        val[0]=0;
        sz=1;
    }
    int idx(char c){
        switch(c){
            case 'A':return 0;
            case 'C':return 1;
            case 'T':return 2;
            case 'G':return 3;
        }
    }
    void insert(char *s,int v){
        int u=0;
        for(int i=0;s[i];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 get_fail(){
        queue<int> q;
        f[0]=0;
        for(int c=0;c<SIGMA_SIZE;c++){
            int u=ch[0][c];
            if(u){
                f[u]=0;
                q.push(u);
            }
        }
        while(!q.empty()){
            int r=q.front();
            q.pop();
            for(int c=0;c<SIGMA_SIZE;c++){
                int u=ch[r][c];
                if(!u){
                    ch[r][c]=ch[f[r]][c];
                    continue;
                }
                q.push(u);
                f[u]=ch[f[r]][c];
                val[u]|=val[f[u]];
            }
        }
    }
}ac;
struct Mat{
    LL mat[MAXNODE][MAXNODE];
    void init(){
        memset(mat,0,sizeof(mat));
    }
}ans;
Mat operator * (Mat a,Mat b){
    int i,j,k,sz=ac.sz;
    Mat ret;
    ret.init();
    for(int k=0;k<sz;k++)
        for(int i=0;i<sz;i++){
            if(!a.mat[i][k]) continue;
            for(int j=0;j<sz;j++) ret.mat[i][j]=(ret.mat[i][j]+a.mat[i][k]*b.mat[k][j])%MOD;
        }
    return ret;
}
Mat operator ^ (Mat a,int n){
    Mat ret,t=a;
    int sz=ac.sz;
    for(int i=0;i<sz;i++)
        for(int j=0;j<sz;j++) ret.mat[i][j]=(i==j);
    while(n){
        if(n&1) ret=ret*t;
        t=t*t;
        n>>=1;
    }
    return ret;
}
void get_mat(){
    for(int u=0;u<ac.sz;u++)
        for(int c=0;c<SIGMA_SIZE;c++) if(!ac.val[u]&&!ac.val[ac.ch[u][c]]) ans.mat[u][ac.ch[u][c]]++;
}
int main(){
    freopen("in.txt","r",stdin);
    while(scanf("%d%d",&M,&N)!=EOF){
        ac.init();
        for(int i=0;i<M;i++){
            scanf("%s",str);
            ac.insert(str,1);
        }
        ac.get_fail();
        ans.init();
        get_mat();
        ans=(ans^N);
        LL cnt=0;
        for(int i=0;i<ac.sz;i++) cnt=(cnt+ans.mat[0][i])%MOD;
        printf("%I64d\n",cnt);
    }
    return 0;
}