POJ 2778 DNA Sequence （AC自动机 + 矩阵快速幂）

题目链接：DNA Sequence

解析：AC自动机 + 矩阵加速（快速幂）。

这个时候AC自动机 的一种状态转移图的思路就很透彻了，AC自动机就是可以确定状态的转移。

AC代码：

#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
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];
    bool end[110];
    int root, L;

    int newnode(){
        for(int i = 0; i < 4; i++) next[L][i] = -1;
        end[L++] = false;
        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;
        else 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] = true;
    }

    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] ] == true) end[now] = true;
            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 ret = Matrix(L);
        for(int i = 0; i < L; i ++)
            for(int j = 0; j < 4; j ++)
                if(end[ next[i][j] ] == false)
                    ret.mat[i][ next[i][j] ] ++;
        return ret;
    }
};

Trie ac;
char buf[20];

Matrix pow_Mat(Matrix a, int n){    //快速幂
    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(){
    #ifdef sxk
        freopen("in.txt", "r", stdin);
    #endif //sxk

    int n, m;
    while(scanf("%d%d", &m, &n) == 2){
        ac.init();
        for(int i = 0; i < m; i ++){
            scanf("%s", buf);
            ac.insert(buf);
        }
        ac.build();
        Matrix a = ac.getMatrix();
        a = pow_Mat(a, n);
        int ans = 0;
        for(int i = 0; i < a.n; i ++)
            ans = (ans + a.mat[0][i]) % MOD;
        printf("%d\n", ans);
    }
    return 0;
}

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