题意

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
    
     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 3330 22 11 00 55 11 44 33 533 0 5 12 1 24 1 2 3 4 */