Work

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. 

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

2

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;
}

hdu 5326 - Work(递归）解题报告

原文：http://blog.csdn.net/luwhere/article/details/47446129