poj 2778 AC自动机与矩阵连乘

http://poj.org/problem?id=2778

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

/**
poj 2778 AC自动机与矩阵连乘
题目大意：给定一些模式串，问可以构造出多少中长度为n且不含模式串中的任何一个作为子串的字符串
解题思路：构造自动机，写出状态转移的矩阵，进行矩阵快速幂，其n次幂就表示长度为n。然后mat[0][i]就表示从根节点到状态点i长度为n的字符串有多少种
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <queue>
using namespace std;
const int MOD=100000;
struct Matrix
{
    int mat[110][110],n;
    Matrix() {}
    Matrix(int _n)
    {
        n=_n;
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<n; j++)
            {
                mat[i][j]=0;
            }
        }
    }
    Matrix operator *(const Matrix &b)const
    {
        Matrix ret=Matrix(n);
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<n; j++)
            {
                for(int k=0; k<n; k++)
                {
                    int tmp=(long long)mat[i][k]*b.mat[k][j]%MOD;
                    ret.mat[i][j]=(ret.mat[i][j]+tmp)%MOD;
                }
            }
        }
        return ret;
    }
};

struct Trie
{
    int next[110][4],fail[110],end[110];
    int root,L;
    int newnode()
    {
        for(int i=0; i<4; i++)
            next[L][i]=-1;
        end[L++]=0;
        return L-1;
    }
    void init()
    {
        L=0;
        root=newnode();
    }
    int getch(char ch)
    {
        if(ch=='A')
            return 0;
        if(ch=='C')
            return 1;
        if(ch=='G')
            return 2;
        return 3;
    }
    void insert(char *s)
    {
        int len=strlen(s);
        int now=root;
        for(int i=0; i<len; i++)
        {
            if(next[now][getch(s[i])]==-1)
                next[now][getch(s[i])]=newnode();
            now=next[now][getch(s[i])];
        }
        end[now]=1;
    }
    void build()
    {
        queue <int>Q;
        for(int i=0; i<4; i++)
        {
            if(next[root][i]==-1)
                next[root][i]=root;
            else
            {
                fail[next[root][i]]=root;
                Q.push(next[root][i]);
            }
        }
        while(!Q.empty())
        {
            int now=Q.front();
            Q.pop();
            if(end[fail[now]]==1)
                end[now]=1;
            for(int i=0; i<4; i++)
            {
                if(next[now][i]==-1)
                    next[now][i]=next[fail[now]][i];
                else
                {
                    fail[next[now][i]]=next[fail[now]][i];
                    Q.push(next[now][i]);
                }
            }
        }
    }
    Matrix getMatrix()
    {
        Matrix res=Matrix(L);
        for(int i=0; i<L; i++)
        {
            for(int j=0; j<4; j++)
            {
                if(end[next[i][j]]==0)
                    res.mat[i][next[i][j]]++;
            }
        }
        return res;
    }
} ac;
char buf[20];

Matrix pow_M(Matrix a,int n)
{
    Matrix ret=Matrix(a.n);
    for(int i=0; i<ret.n; i++)
    {
        ret.mat[i][i]=1;
    }
    Matrix tmp=a;
    while(n)
    {
        if(n&1)ret=ret*tmp;
        tmp=tmp*tmp;
        n>>=1;
    }
    return ret;
}

int main()
{
     int n,m;
     while(~scanf("%d%d",&n,&m))
     {
         ac.init();
         for(int i=0;i<n;i++)
         {
             scanf("%s",buf);
             ac.insert(buf);
         }
         ac.build();
         Matrix a=ac.getMatrix();
         a=pow_M(a,m);
         int ans=0;
         for(int i=0;i<a.n;i++)
         {
             ans=(ans+a.mat[0][i])%MOD;
         }
         printf("%d\n",ans);
     }
     return 0;
}