HDU5036 Explosion(期望 bitset)

题意

题目链接

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<bits/stdc++.h>
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<MAXN> 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;
}