ExplosionTime Limit: 6000/3000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 305 Accepted Submission(s): 86
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
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題意：每個房間都有鑰匙，鑰匙可以打開房間的門，沒有鑰匙的時候只能用炸彈炸門，求所需炸彈的期望。
思路：每個房間需要的炸彈期望爲1/s，s爲聯通這個房間的房間數，最後求和就是答案。
這題讓我認識了bitset，很好用的數據結構。
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <bitset>
using namespace std;
const int inf=0x3f3f3f3f;
const int N=1005;
int t,n,cas=1,a,k;
bitset<N> edge[N];
void init(){
scanf("%d",&n);
for(int i=1;i<=n;i++){
edge[i].reset();
edge[i][i]=1;
}
for(int i=1;i<=n;i++){
scanf("%d",&k);
while(k--){
scanf("%d",&a);
edge[i][a]=1;
}
}
}
void solve(){
init();
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
if(edge[j][i])edge[j]|=edge[i];
}
}
double ans=0;
for(int i=1;i<=n;i++){
int sum=0;
for(int j=1;j<=n;j++){
if(edge[j][i])sum++;
}
ans+=1.0/sum;
}
printf("Case #%d: %.5lf\n",cas++,ans);
}
int main()
{
scanf("%d",&t);
while(t--)solve();
return 0;
}