Work

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

7 2
1 2
1 3
2 4
2 5
3 6
3 7

Sample Output

2

题意：给你n-1个关系，a直接领导b。问这些人中有多少人可以领导K个人。领导包括直接领导和间接领导。

题解：比如1领导2，再去看2能领导几个人，这里就可以想到用递归做。用个计数器记录1能领导（包括间接）的人数，记录下所有人能领导的人数，遍历一遍就能求出答案。

参考代码：

#include<stdio.h>
#include<string.h>
struct node
{
	int num,s[105];
}res[105];
int ans;
void judge(int i)
{
	if(res[i].num==0)
	    return ;
	ans+=res[i].num;
	for(int j=0;j<res[i].num;j++)
	    judge(res[i].s[j]);
}
int main()
{
	int n,k,a,b,cnt;
	while(~scanf("%d%d",&n,&k))
	{
		memset(res,0,sizeof(res));
		for(int i=0;i<n-1;i++)
		{
			scanf("%d%d",&a,&b);
			res[a].s[res[a].num]=b;
			res[a].num++;
		}
		cnt=0;
		for(int i=1;i<=n;i++)
		{
			ans=res[i].num;
			for(int j=0;j<res[i].num;j++)
			    judge(res[i].s[j]);
            if(ans==k)
                cnt++;
		}
		printf("%d\n",cnt);
	}
	return 0;
}

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