(斯特林数)Examining the Rooms--HDOJ

Examining the Rooms

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
The 35th ACM/ICPC Asia Regional Tianjin Site —— Online Contest

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lcy

总结：
斯特林数的应用，根据模板打印出S数组；
S(N,K) 就代表，N个元素，形成K个非空循环的种类数
什么是非空循环？意思就是说，打开一个门，拿到一把钥匙，这个钥匙可以打开另外一个门

因为题目要求砸开的门数量不超过K，那么N个元素里面可以有1个、2个、3个…K个环，即S[N][1] 、S[N][2] 、….S[N][K] 把这些加起来

但是还要考虑到第一个门不能砸开，对应的我们要减去S[N-1][K-1],即除去第一个房间，其他N-1个房间形成的K-1个循环的数目，此时，第一房间独立成为一个循环

总共的排列方式就是 N的阶乘

#include<iostream>#include<stdio.h>#include<algorithm>#include<stack>using namespace std;long long s[30][30];long long fac[30];void cal(){    fac[1] = 1;    for(long long i=2;i<=20 ;++i)        fac[i] = fac[i-1] *i;    for(long long i=1;i<=20;i++)    {        s[i][0]=0;        s[i][i]=1;       for(long long j=1;j<i;j++)           s[i][j]=s[i-1][j-1]+(i-1)*s[i-1][j];    }}int main(void){    long long ncase;    cin >> ncase;    cal();    while(ncase--)    {        long long n,k;        long long sum=0;        scanf("%lld %lld",&n,&k);        if(1 == n || 0 == k)        {            printf("0.0000\n");            continue;        }        for(long long i=1; i<=k; ++i)            sum += s[n][i] - s[n-1][i-1];        printf("%.4f\n",sum*1.0/fac[n]);    }    return 0;}