Examining the Rooms - HDU 3625 第一类斯特林数

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Examining the Rooms

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1205 Accepted Submission(s): 727

Problem Description

A murder happened in the hotel. As the best detective in the town, you should examine all the N rooms of the hotel immediately. However, all the doors of the rooms are locked, and the keys are just locked in the rooms, what a trap! You know that there is exactly one key in each room, and all the possible distributions are of equal possibility. For example, if N = 3, there are 6 possible distributions, the possibility of each is 1/6. For convenience, we number the rooms from 1 to N, and the key for Room 1 is numbered Key 1, the key for Room 2 is Key 2, etc.
To examine all the rooms, you have to destroy some doors by force. But you don’t want to destroy too many, so you take the following strategy: At first, you have no keys in hand, so you randomly destroy a locked door, get into the room, examine it and fetch the key in it. Then maybe you can open another room with the new key, examine it and get the second key. Repeat this until you can’t open any new rooms. If there are still rooms un-examined, you have to randomly pick another unopened door to destroy by force, then repeat the procedure above, until all the rooms are examined.
Now you are only allowed to destroy at most K doors by force. What’s more, there lives a Very Important Person in Room 1. You are not allowed to destroy the doors of Room 1, that is, the only way to examine Room 1 is opening it with the corresponding key. You want to know what is the possibility of that you can examine all the rooms finally.

Input

The first line of the input contains an integer T (T ≤ 200), indicating the number of test cases. Then T cases follow. Each case contains a line with two numbers N and K. (1 < N ≤ 20, 1 ≤ K < N)

Output

Output one line for each case, indicating the corresponding possibility. Four digits after decimal point are preserved by rounding.

Sample Input

Sample Output

0.3333
0.6667
0.6250

题意：有n个房间，其中每个房间都有1把某一个房间钥匙，你最多强行打开m个房间，并且不能强行打开第一个房间。问你能够打开所有房间的概率是多少。

思路：因为每个房间都有一把钥匙，所以可以把其变成几个环，对于每一个环，只要强行打开一种的一个房间即可打开这个环中所有的房间。这里就用到了第一类斯特林数，num[i][j]表示i个房间形成j个环的情况数目。分母为n个房间形成1-n个环的情况数，分子为n个房间形成1-m个环的情况数，减去其中1号房间单独一个环的情况数。

AC代码如下：

#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
int T,t,n,m;
ll num[25][25];
void init()
{
    int i,j,k;
    num[1][1]=1;
    for(i=2;i<=20;i++)
       for(j=1;j<=i;j++)
          num[i][j]=num[i-1][j]*(i-1)+num[i-1][j-1];
}
int main()
{
    int i,j,k;
    ll S=1,a,b;
    init();
    scanf("%d",&T);
    for(t=1;t<=T;t++)
    {
        a=b=0;
        scanf("%d%d",&n,&m);
        for(i=1;i<=n;i++)
           b+=num[n][i];
        for(i=1;i<=m;i++)
           a+=num[n][i];
        for(i=1;i<=m;i++)
           a-=num[n-1][i-1];
        printf("%.4f\n",1.0*a/b);
    }
}