Work

Problem Description

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

Sample Output

2

一个公司管理关系父节点管理子节点以及子节点的子节点以此类推问管理k人数的人有几个

#include<stdio.h>jjjjjjjjjjjjjjjj
#include<cstring>
int pre[1001];
int erzi[10014];
void find(int p)
{
	while(p!=pre[p])
	{
		 p=pre[p];
		 erzi[p]++;
	} 
}
int main()
{
	int n,a,b,m;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(int i=1;i<=n;i++)
		{
			pre[i]=i;
			erzi[i]=0;
		}
		int k=n-1;
		while(k--) 
		{
			scanf("%d%d",&a,&b);
		    pre[b]=a;
		}
		int sum=0;
		for(int i=1;i<=n;i++)
		{
			find(i);
		}
	   for(int i=1;i<=n;i++)
	   {
		if(erzi[i]==m)
		sum++;
	   }
			printf("%d
",sum);
	}
	return 0;
}