【POJ2778】 DNA Sequence AC自动机+矩阵快速幂

DNA Sequence
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 14224 Accepted: 5486

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

题意：给你n个病毒，求m长度的正常DNA序列的个数；
思路：在trie树上的递推，二维数组表示路径的方案数，快速幂解；
从0的dp[0][i]的和；

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<queue>
#include<string.h>
using namespace std;
long long  b[110][110];
int n,m;
char s[115];
int MOD=100000;
int  ans=0;
int flag[110];
int tot;
int nex[110];
long long   cc[110][110];
long long  ccc[110][110];
struct node{
        int ch[4];
}t[110];
int val(char s)
{
    if(s=='A') return 0;
    if(s=='C') return 1;
    if(s=='T') return 2;
    if(s=='G') return 3;
}
inline void insert(char s[])
{
// cout<<" "<<s[0]<<endl;
// cout<<tot<<endl;
    int now=0;
        for(int i=0;s[i];i++)
        {
        int ii=val(s[i]);
//cout<<i<<endl;
                if(!t[now].ch[ii])
                {
                        t[now].ch[ii]=++tot;
                }
                now=t[now].ch[ii];
        }
  flag[now]=1;
}
queue<int> q;
inline void AC()
{
        for(int i=0;i<4;i++) if(t[0].ch[i]) q.push(t[0].ch[i]);
        while(!q.empty())
        {
                int now=q.front();
                q.pop();
                int next=nex[now];
                for(int i=0;i<4;i++)
                {
        // if(now==3) cout<<t[3].ch[2]<<endl;
        // if(now==4) cout<<" "<<t[now].ch[i]<<endl;
                        if(t[now].ch[i])
                        {   
                                nex[t[now].ch[i]]=t[next].ch[i];
                                flag[t[now].ch[i]]|=flag[t[next].ch[i]];
                q.push(t[now].ch[i]);
                        }
                        else t[now].ch[i]=t[next].ch[i];
                }
        }
}
inline void ma_cheng(long long a[110][110],long long  tmp[110][110])
{
        for(int i=0;i<=tot;i++)
        for(int j=0;j<=tot;j++)
    {
        cc[i][j]=0;
        for(int k=0;k<=tot;k++)
            {
                    cc[i][j]+=a[i][k]*tmp[k][j];
                    cc[i][j]%=MOD;
            }

    }

// for(int i=0;i<=5;i++)
    //{
    //for(int j=0;j<=5;j++)
    //cout<<cc[i][j]<<" ";
    //cout<<endl;
    //}
//cout<<endl;

        for(int i=0;i<=tot;i++)
        for(int j=0;j<=tot;j++)
    {
        tmp[i][j]=cc[i][j];
    }
}
inline void ma_mi(int x)
{
    for(int i=0;i<=tot;i++)
    for(int j=0;j<=tot;j++)
    ccc[i][j]=b[i][j];
        while(x>0)
        {
// cout<<" "<<x<<endl;
                if(x&1) ma_cheng(ccc,b);
                ma_cheng(ccc,ccc);
                x>>=1;
        }
}
int main()
{
        scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
        {
                scanf("%s",s);
                insert(s);
        }
        AC();
    //cout<<tot<<endl;
        for(int i=0;i<=tot;i++)
        if(!flag[i])
        {
                for(int j=0;j<4;j++)
                if(!flag[t[i].ch[j]])
                {
                        b[i][t[i].ch[j]]++;
                }
        }
// cout<<tot<<endl;
    //for(int i=0;i<=7;i++)
//{

    //for(int j=0;j<=7;j++)
    //cout<<b[i][j]<<" ";
//cout<<endl;
//}
        ma_mi(m-1);
    ans=0;
        for(int i=0;i<=tot;i++)
        if(!flag[i])
        {
                ans+=b[0][i];
        ans%=MOD;
        }
        cout<<ans<<endl;
}