题意

Sol

和cf上的一道题几乎一摸一样

首先根据期望的线性性，可以转化为求每个点的期望打开次数，又因为每个点最多会被打开一次，只要算每个点被打开的概率就行了

设\(anc[i]\)表示\(i\)的反图中能到达的点集大小，答案等于\(\sum_{i = 1}^n \frac{1}{anc[i]}\)(也就是要保证是第一个被选的)

#include
    
     using namespace std;const int MAXN = 1001;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;bitset
     
       f[MAXN], Empty;double solve() {    N = read();    for(int i = 1; i <= N; i++) f[i] = Empty;    for(int i = 1; i <= N; i++) {        int k = read();  f[i].set(i);        for(int j = 1; j <= k; j++) {            int v = read(); f[v].set(i);        }    }    for(int k = 1; k <= N; k++)         for(int i = 1; i <= N; i++)             if(f[i][k]) f[i] = f[i] | f[k];    double ans = 0;    for(int i = 1; i <= N; i++) ans += 1.0 / f[i].count();    return ans;}int main() {    int T = read();    for(int i = 1; i <= T; i++) printf("Case #%d: %.5lf\n", i, solve());    return 0;}