hdoj 5326 Work【并查集】

最新推荐文章于 2017-12-13 12:21:41 发布

原创最新推荐文章于 2017-12-13 12:21:41 发布 · 334 阅读

0 ·

CC 4.0 BY-SA版权

并查集专栏收录该内容

15 篇文章

订阅专栏

本文探讨了一个有趣的问题：在一个公司的组织结构中，如何计算特定数量的员工由多少位管理者直接或间接领导。通过构建公司内部的领导关系树，并利用递归算法进行遍历，实现了对每位员工下属数量的有效统计。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Work

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1711 Accepted Submission(s): 1027

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

代码：

#include<cstdio>
int per[110],num[110];//记录下属的个数 
void find(int x)
{
	if(per[x]!=x)
	{
		num[per[x]]++;	//这里主要是运用了查找时候的递归因素，逐个往上递归找根结点的时候，把每个上级所带领的下属个数都加1 
		find(per[x]);
		return ;
	 } 	
}
 int main()
 {
 	int n,k;
 	while(~scanf("%d%d",&n,&k))
 	{
 		for(int i=1;i<=n;i++)
 		{
 			per[i]=i;
 			num[i]=0;
		 }
		 for(int i=1;i<n;i++)
		 {
		 	int a,b;
		 	scanf("%d%d",&a,&b);
			if(a!=b)
				per[b]=a;
		 }
		 for(int i=1;i<=n;i++)//寻找每个点的下属的个数 
		 	find(i);
		 int ans=0;
		for(int i=1;i<=n;i++)
		{
			if(num[i]==k)
				ans++;
		}
		 printf("%d\n",ans);
	 }
	 return 0;
 }