poj 2278 DNASequnce AC自动机

地址：http://poj.org/problem?id=2778

题目：

DNA Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15453		Accepted: 5964

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

思路： 用病毒源建立ac自动机，然后所有节点可看做状态转移图。呃，懒得画图了，你们可以百度下。然后根据转移图建立矩阵，之后快速幂即可。
　　之前一直忘了把end【x】为1的节点的所有子节点的end标记为1，wa的不省人事。

#include <queue>
#include <cstring>
#include <cstdio>
using namespace std;

const long long mod=1e5;
struct MM
{
    int r,c;
    long long mx[105][105];
    MM(int rr=0,int cc=0){r=rr,c=cc;}
    friend MM operator *(MM ta,MM tb)
    {
        MM tc(ta.r,tb.c);
        for(int i=0;i<tc.r;i++)
            for(int j=0;j<tc.c;j++)
            {
                tc.mx[i][j]=0;
                for(int k=0;k<tb.r;k++)
                    tc.mx[i][j]=(tc.mx[i][j]+ta.mx[i][k]*tb.mx[k][j])%mod;
            }
        return tc;
    }
    friend MM operator ^(MM ta,int num)
    {
        MM ret(ta.r,ta.c);                          //r==c
        memset(ret.mx,0,sizeof ret.mx);
        for(int i=0;i<ta.r;i++) ret.mx[i][i]=1;    //µ¥Î»¾ØÕó
        while(num)
        {
            if(num&1)
                ret=ta*ret;
            num>>=1;
            ta=ta*ta;
        }
        return ret;
    }
}mm;
struct AC_auto
{
    const static int LetterSize = 4;
    const static int TrieSize = 4*1010;

    int tot,root,fail[TrieSize],end[TrieSize],next[TrieSize][LetterSize];

    int newnode(void)
    {
        memset(next[tot],-1,sizeof(next[tot]));
        end[tot] = 0;
        return tot++;
    }

    void init(void)
    {
        tot = 0;
        root = newnode();
    }

    int getidx(char x)
    {
        if(x=='A')return 0;
        else if(x=='C')return 1;
        else if(x=='G')return 2;
        else return 3;
    }

    void insert(char *ss)
    {
        int len = strlen(ss);
        int now = root;
        for(int i = 0; i < len; i++)
        {
            int idx = getidx(ss[i]);
            if(next[now][idx] == -1)
                next[now][idx] = newnode();
            now = next[now][idx];
        }
        end[now]=1;
    }

    void build(void)
    {
        queue<int>Q;
        fail[root] = root;
        for(int i = 0; i < LetterSize; i++)
            if(next[root][i] == -1)
                next[root][i] = root;
            else
                fail[next[root][i]] = root,Q.push(next[root][i]);
        while(Q.size())
        {
            int now = Q.front();Q.pop();
            if(end[fail[now]])end[now]=1;
            for(int i = 0; i < LetterSize; 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]);
        }
    }

    int buildmm(int n)
    {
        int cnt = 0;
        mm.c=mm.r=tot;
        for(int i = 0; i < tot; i++)
        for(int j = 0; j < 4; j++)
        if(!end[next[i][j]])
            mm.mx[i][next[i][j]]++;
        mm=mm ^ n;
        for(int i = 0; i < tot; i++)
            cnt = (cnt + mm.mx[0][i]) % mod;
        return cnt;
    }
};


AC_auto ac;
char sa[20];
int main(void)
{
    int n,m;
    scanf("%d%d",&n,&m);
    ac.init();
    for(int i=1;i<=n;i++)
        scanf("%s",sa),ac.insert(sa);
    ac.build();
    printf("%d
",ac.buildmm(m));
    return 0;
}