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Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1781    Accepted Submission(s): 1065

Problem Description
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It’s an interesting experience to move from ICPC to work, end my college life and start a brand new journey in company.
As is known to all, every stuff in a company has a title, everyone except the boss has a direct leader, and all the relationship forms a tree. If A’s title is higher than B(A is the direct or indirect leader of B), we call it A manages B.
Now, give you the relation of a company, can you calculate how many people manage k people.
 

Input
There are multiple test cases.
Each test case begins with two integers n and k, n indicates the number of stuff of the company.
Each of the following n-1 lines has two integers A and B, means A is the direct leader of B.

1 <= n <= 100 , 0 <= k < n
1 <= A, B <= n
 

Output
For each test case, output the answer as described above.
 

Sample Input
7 2 1 2 1 3 2 4 2 5 3 6 3 7
 

Sample Output
2
 

Author
ZSTU
 

Source
2015 Multi-University Training Contest 3 

思路：
题意就是，给了你每组a,b代表a是b的祖先，且具有传递性，那么问你有几个人，它的儿子个数是m。思想就是类似于并查集找祖先的思想，对于每个点，都去找它的祖先路径，并且没经过一个点那么这个点的父亲数++并且经过的点的儿子数++，直到遇到一个点它的祖先是它自己为止。然后对这些数遍历，就得到了每个点的父亲数和儿子数。


代码： 
#include<iostream>
#include<stdio.h>
#include<cstring> 
using namespace std;
 
int ancestor[1001];  //储存祖先 
int son[10014];  //储存儿子 
        
void find(int p)  
{  
    while(p!=ancestor[p])  
    {  
         p=ancestor[p];  
         son[p]++;  
    }   
}  

int main()  
{  
    int n,a,b,m;  
    while(scanf("%d%d",&n,&m)!=EOF)  
    {  
        for(int i=1;i<=n;i++)  
        {  
            ancestor[i]=i;  
            son[i]=0;  
        }  
        for(int i=1;i<n;i++) 
        {  
            scanf("%d%d",&a,&b);  
            ancestor[b]=a;  
        }  
        int sum=0;  
        for(int i=1;i<=n;i++)  
        {  
            find(i);  
        }  
       for(int i=1;i<=n;i++)  
       {  
            if(son[i]==m)  
            sum++;  
       }  
            printf("%d\n",sum);  
    }  
    return 0;  
}