【杭电oj】3625 - Examining the Rooms（第一类斯特林数打表）

点击打开题目

Examining the Rooms

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1437 Accepted Submission(s): 873

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

   3
3 1
3 2
4 2

Sample Output

   0.3333
0.6667
0.6250

   
    
     Hint
    
Sample Explanation

When N = 3, there are 6 possible distributions of keys:

	Room 1	Room 2	Room 3	Destroy Times
#1	Key 1	Key 2	Key 3	Impossible
#2	Key 1	Key 3	Key 2	Impossible
#3	Key 2	Key 1	Key 3	Two
#4	Key 3	Key 2	Key 1	Two
#5	Key 2	Key 3	Key 1	One
#6	Key 3	Key 1	Key 2	One

In the first two distributions, because Key 1 is locked in Room 1 itself and you can’t destroy Room 1, it is impossible to open Room 1. 
In the third and forth distributions, you have to destroy Room 2 and 3 both. In the last two distributions, you only need to destroy one of Room 2 or Room 

Source

2010 Asia Regional Tianjin Site —— Online Contest

第一类斯特林数表示：n和节点构成非空k个环的次数。

而n个钥匙放在n个房间的个数用排列公式：N！

然后就是处理1不能单独成环的问题了：S[ n ] [ k ] = S[ n ] [ k ] - S [ n - 1 ] [ k - 1 ] 意义为：减去n-1个节点时形成k-1个环的个数（多的那一个即为1单独成的环）

最后算一下比值就行了。

代码如下：

#include <cstdio>
double f[22]={1};
__int64 S[22][22];
void S1()
{
	S[1][1] = 1;
	S[1][0] = 0;
	for (int i = 1 ; i <= 20 ; i++)
	{
		S[i][0] = 0;
		S[i][i] = 1;
	}
	for (int i = 2 ; i <= 20 ; i++)
		for (int j = 1 ; j < i ; j++)
			S[i][j] = (i - 1) * S[i-1][j] + S[i-1][j-1];
}
int main()
{
	S1();		//求第一类斯特林数 
	for (int i = 1 ; i <= 20 ; i++)		//排列公式求阶乘 
		f[i] = f[i-1] * i;
	int u;
	int n,k;
	scanf ("%d",&u);
	while (u--)
	{
		scanf ("%d %d",&n,&k);
		__int64 sum = 0;
		for (int i = 1 ; i <= k ; i++)
			sum += S[n][i] - S[n-1][i-1];		//把1单独成环排除 
		printf ("%.4lf\n",(double)sum / f[n]);
	}
	return 0;
}