POJ 2778（自动机+矩阵幂乘（字符串果然是个坑爹玩应！））

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

关键是状态转移，矩阵就是解决状态转移的！（比较难）

数据量：1 <= n <=2000000000（明显快速算法！）

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

//#pragma comment(linker, "/STACK:102400000")
#include<cstdlib>
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<set>
#include<map>
#include<list>
#include<queue>
#include<stack>
#include<vector>
#define tree int o,int l,int r
#define lson o<<1,l,mid
#define rson o<<1|1,mid+1,r
#define lo o<<1
#define ro o<<1|1
#define pb push_back
#define mp make_pair
#define ULL unsigned long long
#define LL long long
#define inf 0x7fffffff
#define eps 1e-7
#define N 109
#define M 4
#define mod 100000
using namespace std;
int m,n,T,t,x,y,u;
int ch[N][4];
int v[N];
int f[N],last[N],num;
LL ma[N][N],temp[N][N];
void clear()//Trie树初始化
{
    memset(ma,0,sizeof(ma));
    num=1;
    memset(ch[0],0,sizeof(ch[0]));
    memset(v,0,sizeof(v));
    memset(last,0,sizeof(last));
}
int idx(char c)
{
    switch(c)
    {
        case 'A':return 0;
        case 'C':return 1;
        case 'G':return 2;
        case 'T':return 3;
    }
    return 0;
}
void insert(char str[],int value)//建Trie树
{
    int len=strlen(str);
    int u=0;
    for (int i=0; i<len; ++i )
    {
        int c=idx(str[i]);
        if(!ch[u][c])//保存的是结点坐标
        {
            memset(ch[num],0,sizeof(ch[num]));
            ch[u][c]=num++;//
        }
        u=ch[u][c];
    }
    v[u]=value;
}
void getac()
{
    queue<int> q;//保存的节点下标
    f[0]=0;
    for (int c=0; c<M; ++c )
    {
        int u=ch[0][c];
        if(u)//不需要优化的else
        {
            q.push(u);
            f[u]=0;
            last[u]=v[u];//WA,可能有长度为1的串
        }
    }
    while(!q.empty())
    {
        int r=q.front();
        q.pop();
        for (int c=0; c<M; ++c )
        {
            int u=ch[r][c];
            if(u)
            {
                q.push(u);
                int s=f[r];
                f[u]=ch[s][c];
                last[u]=(v[u]||last[f[u]]);//改
            }
            else //重要优化
                ch[r][c]=ch[f[r]][c];
        }
    }
}
char str[20];
void build()//构造矩阵！！！
{
    for(int i=0;i<num;i++)
    {
        for(int j=0;j<4;j++)
        {
            int u=ch[i][j];
            if(last[u]==0)
            ma[i][u]++;
        }
    }
}
void malti(LL a[][N],LL b[][N],int n)
{
    LL c[N][N]={0};
    for(int i=0;i<n;i++)
    for(int j=0;j<n;j++)
    for(int k=0;k<n;k++)
    {
        c[i][j]+=a[i][k]*b[k][j];
        c[i][j]%=mod;
    }
    memcpy(a,c,sizeof(c));
}
void mat(LL ma[][N],int n,int num)
{
    memset(temp,0,sizeof(temp));
    for(int i=0;i<num;i++)
    temp[i][i]=1;
    while(n)
    {
        if(n&1)
        {
            malti(temp,ma,num);
        }
        malti(ma,ma,num);
        n>>=1;
    }
    memcpy(ma,temp,sizeof(temp));
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("ex.in","r",stdin);
#endif
    int ncase=0;
//    scanf("%d",&T);
//    while(T--)
    while(scanf("%d%d%*c",&m,&n)==2)
    {
        clear();
        while(m--)
        {
            scanf("%s",str);
            insert(str,1);
        }
        getac();
        build();
//        print2(ma,num,num);
        mat(ma,n,num);
        int ans=0;
        for(int i=0;i<num;i++)
        {
            ans+=ma[0][i];
            ans%=mod;
        }
        printf("%d
",ans);
    }
    return 0;
}