POJ 2778 DNA SequenceAC自动机矩阵快速幂-优快云博客

本文链接：https://blog.youkuaiyun.com/ACMer_AK/article/details/52211106

本文介绍了一种结合AC自动机与矩阵快速幂的方法来高效解决字符串匹配问题中的计数任务。通过构建AC自动机并利用矩阵快速幂进行状态转移，可以在O(log N)的时间复杂度内求解经过特定节点的路径数量。

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#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
#include <queue>
#include <math.h>
#define mod 100000
#define LL long long

using namespace std;
const int M = 4;
const int N = 1e6+10;
struct Matrix{
    int mat[110][110];
    int n;
    Matrix(){}
    Matrix(int nn){
        n = nn;
        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 = (LL)mat[i][k]*b.mat[k][j]%mod;
                    ret.mat[i][j]=(ret.mat[i][j]+tmp)%mod;
                }
            }
        }
        return ret;
    }
};
struct Trie{
    int next[N][M],fail[N];
    bool end[N];
    int root,L;
    int getIndex(char ch){
        switch(ch){
        case 'A':
            return 0;
        case 'C':
            return 1;
        case 'G':
            return 2;
        case 'T':
            return 3;
        }
    }
    int newNode(){
        for(int i = 0; i < M; ++i){
            next[L][i] = -1;
        }
        end[L++] = false;
        return L-1;
    }
    void init(){
        L = 0;
        root = newNode();
    }
    void insert(char buf[]){
        int now = root;
        for(int i = 0 ; buf[i]; ++i){
            int x = getIndex(buf[i]);
            if(next[now][x] == -1){
                next[now][x] = newNode();
            }
            now = next[now][x];
        }
        end[now] = true;
    }
    void build(){
        queue<int>Q;
        for(int i = 0; i < M; ++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]] == true)
                end[now] = true;
            for(int i = 0; i < M; ++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]);
                }
            }
        }
    }
    //计算的是选择0个、1个、、、n个的总数目
//    Matrix getMatrix(){
//        Matrix res = Matrix(1+L);
//        for(int i = 0; i < L; ++i){
//            for(int j = 0; j < M; ++j)
//                if(end[next[i][j]] == false)
//                    res.mat[i][next[i][j]]++;
//        }
//        for(int i = 0; i < L+1; ++i){
//            res.mat[i][L] = 1;
//        }
//        return res;
//    }
    //计算选择n个的数目
    Matrix getMatrix(){  
        Matrix res = Matrix(L);  
        for(int i = 0; i < L; ++i){  
            for(int j = 0; j < M; ++j)  
                if(end[next[i][j]] == false)  
                    res.mat[i][next[i][j]]++;  
        }  
        return res;  
    }  
};
Trie ac;
char buf[20];
Matrix pow_Matrix(Matrix a,int n){
    Matrix ret = Matrix(a.n);
    for(int i = 0; i < ret.n; ++i)
        ret.mat[i][i] = 1;
    while(n){
        if(n&1){
            ret = ret*a;
        }
        a = a*a;
        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_Matrix(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;
}