Problem Description Everyone knows Matt enjoys playing games very much. Now, he is playing such a game. There are N rooms, each with one door. There are some keys(could be none) in each room corresponding to some doors among these N doors. Every key can open only one door. Matt has some bombs, each of which can destroy a door. He will uniformly choose a door that can not be opened with the keys in his hand to destroy when there are no doors that can be opened with keys in his hand. Now, he wants to ask you, what is the expected number of bombs he will use to open or destroy all the doors. Rooms are numbered from 1 to N.
Input The first line of the input contains an integer T, denoting the number of testcases. Then T test cases follow.
In the first line of each test case, there is an integer N (N<=1000) indicating the number of rooms.
The following N lines corresponde to the rooms from 1 to N. Each line begins with an integer k (0<=k<=N) indicating the number of keys behind the door. Then k integers follow corresponding to the rooms these keys can open.
Output For each test case, output one line "Case #x: y", where x is the case number (starting from 1), y is the answer which should be rounded to 5 decimal places.
Sample Input
2
3
1 2
1 3
1 1
3
0
0
0
Sample Output
Case #1: 1.00000
Case #2: 3.00000
Source 2014 ACM/ICPC Asia Regional Beijing Online
题意：n个房间，每个房间都有若干个钥匙打开其他的门，如果手上没有钥匙可以选择等概率随机选择一个门炸开，求用炸弹数的期望。
思路：每个房间期望都是可加的。 单独考虑一个房间，如果有k个房间被炸开都会导致这个房间被打开。那么炸一次这个房间被打开的概率即为
kn
。也就意味着为了把这个房间打开的期望炸的次数为
nk
。全部加起来后除以n即可。bitset优化。
#include
#include
#include
#include
#include
using namespace std; const int maxn = 1005; bitset
a[maxn]; int main() { int t, cas = 1; int n; scanf("%d", &t); while (t--) { scanf("%d", &n); for (int i = 0; i < n; i++) { a[i].reset(); a[i][i] = 1; } int c, x; for (int i = 0; i < n; i++) { scanf("%d", &c); while (c--) { scanf("%d", &x); a[i][--x] = 1; } } for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) if (a[j][i]) a[j] |= a[i]; double ans = 0; for (int i = 0; i < n; i++) { c = 0; for (int j = 0; j < n; j++) if (a[j][i]) c++; ans += 1.0 / c; } printf("Case #%d: %.5lf\n",cas++,ans); } return 0; }