bzoj 1898 矩阵快速幂

思路：因为鱼的周期为2, 3, 4, 所以以12个为周期，我们拿走12步得到的矩阵进行快速幂，余下的再进行一次矩阵乘法。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PII pair<int, int>
#define y1 skldjfskldjg
#define y2 skldfjsklejg
using namespace std;

const int N =  + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = ;

int n, m, S, T, K, num;
int fish[][], len[], p[];
int ban[N];

vector<int> edge[N];

struct Matrix {
    int a[][], n;
    Matrix(int _n = ) {
        n = _n;
        memset(a, , sizeof(a));
    }
    void init() {
        for(int i = ; i < n; i++)
            a[i][i] = ;
    }
    Matrix operator * (const Matrix &B) const {
        Matrix C(n);
        for(int i = ; i < n; i++)
            for(int j = ; j < n; j++)
                for(int k = ; k < n; k++)
                    C.a[i][j] = (C.a[i][j] + a[i][k] * B.a[k][j]) % mod;
        return C;
    }
    Matrix operator ^ (int b) {
        Matrix C(n); C.init();
        Matrix A = (*this);
        while(b) {
            if(b & ) C = C * A;
            A = A * A; b >>= ;
        }
        return C;
    }
};

void nextTime() {
    for(int i = ; i < num; i++) {
        ban[fish[i][p[i]]]--;
        p[i] = (p[i] + ) % len[i];
        ban[fish[i][p[i]]]++;
    }
}

int main() {
    scanf("%d%d%d%d%d", &n, &m, &S, &T, &K);
    for(int i = ; i <= m; i++) {
        int u, v; scanf("%d%d", &u, &v);
        edge[u].push_back(v);
        edge[v].push_back(u);
    }

    scanf("%d", &num);
    for(int i = ; i < num; i++) {
        scanf("%d", &len[i]);
        for(int j = ; j < len[i]; j++)
            scanf("%d", &fish[i][j]);
        ban[fish[i][]]++;
    }

    Matrix B[];
    for(int i = ; i < ; i++) B[i].n = n;
    for(int i = ; i < n; i++)
        if(!ban[i]) B[].a[i][i] = ;
    for(int k = ; k <= ; k++) {
        nextTime();
        for(int u = ; u < n; u++) {
            if(ban[u]) continue;
            for(int i = ; i < edge[u].size(); i++) {
                int v = edge[u][i];
                for(int j = ; j < n; j++)
                    B[k].a[u][j] = (B[k].a[u][j] + B[k - ].a[v][j]) % mod;
            }
        }
    }
    Matrix A = B[K % ] * (B[] ^ (K / ));
    printf("%d\n", A.a[T][S]);
    return ;
}

/*
*/