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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.
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.
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.
2 3 1 2 1 3 1 1 3 0 0 0
Case #1: 1.00000 Case #2: 3.00000
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<algorithm> 6 #include<stdlib.h> 7 #include<bitset> 8 using namespace std; 9 #define N 1006 10 int n; 11 bitset<N> bs[N]; 12 int main() 13 { 14 int t; 15 int ac=0; 16 scanf("%d",&t); 17 while(t--){ 18 scanf("%d",&n); 19 for(int i=0;i<N;i++){ 20 bs[i].reset(); 21 bs[i][i]=true; 22 } 23 for(int i=0;i<n;i++){ 24 int k; 25 scanf("%d",&k); 26 for(int j=0;j<k;j++){ 27 int x; 28 scanf("%d",&x); 29 x--; 30 bs[i][x]=true; 31 } 32 } 33 for(int j=0;j<n;j++){ 34 for(int i=0;i<n;i++){ 35 if(bs[i][j]){ 36 bs[i] |= bs[j]; 37 } 38 } 39 } 40 double ans=0; 41 for(int j=0;j<n;j++){ 42 int cnt=0; 43 for(int i=0;i<n;i++){ 44 if(bs[i][j]){ 45 cnt++; 46 } 47 } 48 ans+=1.0/cnt; 49 } 50 printf("Case #%d: %.5lf\n",++ac,ans); 51 } 52 return 0; 53 }
hdu 5036 Explosion（概率期望+bitset）
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原文地址：http://www.cnblogs.com/UniqueColor/p/4790034.html
