POJ2278 DNA Sequence —— AC自动机 + 矩阵优化

题目链接：https://vjudge.net/problem/POJ-2778

 

DNA Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18479		Accepted: 7112

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

Source

POJ Monthly--2006.03.26,dodo

 

 

题意：

给出m个DNA序列，问长度为n且不含这m个序列的DNA有多少个？

 

题解：

1.把这m个序列插入到AC自动机中。

2.根据自动机中各个状态之间的关系，构成一张邻接矩阵A，但需要去除与“结束点”有关的边，这样就能保证不含有给出的序列。

3.长度为n，那么答案就是 A^n 中，初始状态那一行之和。

 

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const double EPS = 1e-6;
const int INF = 2e9;
const LL LNF = 9e18;
const int MOD = 1e5;
const int MAXN = 110+10;

int Size;
int Map[128];
struct MA
{
    int mat[110][110];
    void init()
    {
        for(int i = 0; i<Size; i++)
        for(int j = 0; j<Size; j++)
            mat[i][j] = (i==j);
    }
};

MA operator*(const MA &x, const MA &y)
{
    MA ret;
    memset(ret.mat, 0, sizeof(ret.mat));
    for(int i = 0; i<Size; i++)
    for(int j = 0; j<Size; j++)
    for(int k = 0; k<Size; k++)
        ret.mat[i][j] += (1LL*x.mat[i][k]*y.mat[k][j])%MOD, ret.mat[i][j] %= MOD;
    return ret;
}

MA qpow(MA x, int y)
{
    MA s;
    s.init();
    while(y)
    {
        if(y&1) s = s*x;
        x = x*x;
        y >>= 1;
    }
    return s;
}

struct Trie
{
    const static int sz = 4, base = 'A';
    int next[MAXN][sz], fail[MAXN], end[MAXN];
    int root, L;
    int newnode()
    {
        for(int i = 0; i<sz; i++)
            next[L][i] = -1;
        end[L++] = false;
        return L-1;
    }
    void init()
    {
        L = 0;
        root = newnode();
    }
    void insert(char buf[])
    {
        int len = strlen(buf);
        int now = root;
        for(int i = 0; i<len; i++)
        {
            if(next[now][Map[buf[i]]] == -1) next[now][Map[buf[i]]] = newnode();
            now = next[now][Map[buf[i]]];
        }
        end[now] = true;
    }
    void build()
    {
        queue<int>Q;
        fail[root] = root;
        for(int i = 0; i<sz; 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();
            end[now] |= end[fail[now]]; //当前串的后缀是否也包含单词
            for(int i = 0; i<sz; 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 query(int n)
    {
        MA s;
        memset(s.mat, 0, sizeof(s.mat));
        for(int i = 0; i<L; i++)
        {
            if(end[i]) continue;    //存在单词的状态没有后继
            for(int j = 0; j<sz; j++)
            {
                if(end[next[i][j]]) continue;   //存在单词的状态没有前驱
                s.mat[i][next[i][j]]++; // i到next[i][j]的路径数+1。注意，当next[i][j]==root时，路径数很可能大于1。
            }
        }

        int ret = 0;
        Size = L;
        s = qpow(s, n);
        for(int i = 0; i<L; i++)    //答案为：初始状态到各个状态（包括初始状态）的路径数之和。
            ret = (ret+s.mat[0][i])%MOD;
        return ret;
    }
};

Trie ac;
char buf[20];
int main()
{
    Map['A'] = 0; Map['C'] = 1; Map['G'] = 2; Map['T'] = 3; //离散化
    int n, m;
    while(scanf("%d%d", &m,&n)!=EOF)
    {
        ac.init();
        for(int i = 1; i<=m; i++)
        {
            scanf("%s", buf);
            ac.insert(buf);
        }
        ac.build();
        int ans = ac.query(n);
        printf("%d\n", ans);
    }
    return 0;
}