HDU3625 Examining the Rooms

@(HDU)[Stirling數]

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

Solution

題意:

n个房间对应n把钥匙, 每个房间的钥匙随机放在某个房间内, 概率相同。
有K次炸门的机会，求能进入所有房间的概率
一号门不给你炸
--by ZWL

實際上這題就是第一類Stirling數的模板應用.
不妨設(key_i)表示第(i)個房間內放的鑰匙是哪把, 則對於這樣一組(i)和(key_i), 可看作在一個有向圖中, 點(i)向點(key_i)連出一條有向邊. 當對這個有向圖連邊完畢后, 就會發現, 每個點的出度和入度都為(1). 連邊組成一個或多個環. 而在一個環中, 只要有一個房間可以通過任何方式進入, 則其他房間都可以進入了.
由於可以炸開門的次數為(k), 因此要求這個圖中環的個數不超過(k). 同時由於(1)號房間的門不能被炸開, 因此(1)不能單獨在一個環中. 所以滿足條件的方案數為: $$qua = sum_{i = 1}^n left( left[ egin{array}{} n i end{array}{} ight] - left[ egin{array}{} n - 1 i - 1 end{array}{} ight] ight)$$
代碼附上:

#include<cstdio>
#include<cctype>
using namespace std;

inline int read()
{
	int x = 0, flag = 1;
	char c;
	while(! isdigit(c = getchar()))
		if(c == '-')
			flag *= - 1;
	while(isdigit(c))
		x = x * 10 + c - '0', c = getchar();
	return x * flag;
}

const int N = 1 << 5;

long long f[N][N];
long long fac[N];

int main()
{
	fac[0] = 1;
	for(int i = 1; i < N; i ++)
		fac[i] = fac[i - 1] * i;
	
	f[0][0] = 1;
	
	for(int i = 1; i < N; i ++)
		f[0][i] = 0;
		
	for(int i = 1; i < N; i ++)
	{
		f[i][0] = (long long)0;
		
		for(int j = 1; j <= i; j ++)
			f[i][j] = f[i - 1][j - 1] + (long long)(i - 1) * f[i - 1][j];
	}
	
	int T = read();
	
	for(; T --; )
	{
		int n = read(), k = read();
		
		long long qua = 0;
		
		for(int i = 1; i <= k; i ++)
			qua += f[n][i] - f[n - 1][i - 1];
			
		printf("%.4lf
", (double)qua / fac[n]);
	}
}