BZOJ1898: [Zjoi2005]Swamp 沼泽鳄鱼(矩阵快速幂)

题意

题目链接

Sol

不难发现吃人鱼的运动每\(12s\)一个周期

所以暴力建12个矩阵，放在一起快速幂即可

最后余下的部分暴力乘

#include<bits/stdc++.h>

using namespace std;

const int MAXN = 52, mod = 10000;

inline int read() {

    char c = getchar(); int x = 0, f = 1;

    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}

    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();

    return x * f;

}

int N, M, S, T, K, mp[MAXN][MAXN], pos[MAXN];

int add(int x, int y) {

    if(x + y < 0) return x + y + mod;

    return x + y >= mod ? x + y - mod : x + y;

}

int add2(int &x, int y) {

    if(x + y < 0) x = x + y + mod;

    else x = (x + y >= mod ? x + y - mod : x + y);

}

int mul(int x, int y) {

    return 1ll * x * y % mod;

}

struct Ma {

    int m[MAXN][MAXN];

    Ma() {

        memset(m, 0, sizeof(m));

    }

    Ma operator * (const Ma &rhs) const {

        Ma ans;

        for(int k = 1; k <= N; k++)

            for(int i = 1; i <= N; i++)

                for(int j = 1; j <= N; j++)

                    add2(ans.m[i][j], mul(m[i][k], rhs.m[k][j]));

        return ans;

    }

}a[15], bg;

Ma MatrixFp(Ma a, int p) {

    Ma base;

    for(int i = 1; i <= N; i++) base.m[i][i] = 1;

    while(p) {

        if(p & 1) base = base * a;

        a = a * a; p >>= 1;

    }

    return base;

}

int main() {

    N = read(); M = read(); S = read() + 1; T = read() + 1; K = read();

    for(int i = 1; i <= M; i++) {int x = read() + 1, y = read() + 1; bg.m[x][y]++; bg.m[y][x]++;}

    int fn = read();

    for(int i = 1; i <= 12; i++) a[i] = bg;

    for(int i = 1; i <= fn; i++) {

        int num = read();

        for(int j = 1; j <= num; j++) pos[j] = read() + 1;

        for(int j = 1; j <= 12; j++)

            for(int k = 1; k <= N; k++)

                a[j].m[k][pos[j % num + 1]] = 0;

    }

    Ma res = a[1];

    //for(int i = 1; i <= N; i++) res.m[i][i] = 1;

    for(int i = 2; i <= 12; i++) res = res * a[i];

    res = MatrixFp(res, K / 12);

    for(int i = 1; i <= K % 12; i++) res = res * a[i];

    printf("%d\n", res.m[S][T]);

    return 0;

}

/*

6 8 1 5 333

0 2

2 1

1 0

0 5

5 1

1 4

4 3

3 5

3

3 0 5 1

2 1 2

4 1 2 3 4

*/