杭电 5326 Work （并查集求子结点为k的结点数）

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

大意：
　　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。
思路：
　　用数组num[]记录每个结点领导的人数，令i从1~n表示结点，若i不是父结点，让num[fa[i]]+1,表示i上面的结点领导的人数加1，一直加到父结点，到i=n时sun[1~n]表示1~n个结点领导的人数，再找出人数等于k的。

#include<cstdio>
#include<string.h>
#include<algorithm>
using namespace std;
int n,k,i,ans,fa[110],a,b,num[110];
void f1(int a)                        //若a不是父结点，让a上面所有结点都加1；表示领导的人数 
{
    while(a != fa[a])
    {
        num[fa[a]]++;
        a=fa[a];
    }
}
int main()
{
    while(scanf("%d %d",&n,&k)!=EOF)
    {
        memset(num,0,sizeof(num));
        ans=0;
        for(i = 1 ; i <= n ; i++)
        {
            fa[i]=i;
        }
        for(i = 1 ; i < n ; i++)
        {
            scanf("%d %d",&a,&b);
            fa[b]=a;
        }
        for(i = 1 ; i <= n ; i++)
        {
            f1(i);
        }
        for(i = 1; i<=n;i++)
        {
            if(num[i] == k)                //找出领导的人数为k的结点 
            {
                ans++;
            }
        }
        printf("%d
",ans);
    }
}