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
*/