HDOJ 5326 Work (并查集)

Work

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1201    Accepted Submission(s): 734

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
   
   
   
   

 

题目大意：有n个人，各自管理不同的人，问有多少人管理k个人。

思路：先记录每个人的直接上级，然后模拟寻根路径压缩的过程进行计数求和。

ac代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#define MAXN 2555
#define mod 1000000007
#define INF 0xfffffff
#define MIN(a,b) a>b?b:a
using namespace std;
int sum[MAXN];
int pri[MAXN];
int find(int x)
{
	int r=x;
	while(r!=pri[r])
	r=pri[r];
	int i=x,j;
	while(i!=r)
	{
		j=pri[i];
		sum[j]++;//并没有压缩路径，只是把那条路径上的结点的sum值全都加1
		i=j;
	}
	return r;
}
int main()
{
	int n,i,k,a,b;
	while(scanf("%d%d",&n,&k)!=EOF)
	{
		for(i=1;i<=n;i++)
		pri[i]=i;
		memset(sum,0,sizeof(sum));
		for(i=1;i<n;i++)
		{
			scanf("%d%d",&a,&b);
			if(a!=b)//不用寻根，直接连
			pri[b]=a;
		}
		for(i=1;i<=n;i++)
		find(i);
		int ans=0;
		for(i=1;i<=n;i++)
		{
			if(sum[i]==k)
			ans++;
		}
		printf("%d\n",ans);
	}
	return 0;
}