hdoj Work

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                                        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.

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 21 21 32 42 53 63 7

Sample Output

2

题解：这是并查集的题目，大意就是找领导者和员工之间的关系

下面是我的代码：

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;const int MAX=1e3+10;int par[MAX],num[MAX];void find(int m){if(par[m]!=m){num[par[m]]++;//记录 每个上级所带领的下属个数 find(par[m]);return ;}}void un(int x,int y){if(x!=y)par[y]=x;}int main(){int n,m,ans;while(scanf("%d%d",&n,&m)!=EOF){int a,b;memset(num,0,sizeof(num));for(int i=1;i<=n;i++)par[i]=i;for(int j=1;j<n;j++){scanf("%d%d",&a,&b);un(a,b);}for(int i=1;i<=n;i++)find(i);//寻找每个点的下属个数 ans=0;for(int i=1;i<=n;i++){if(num[i]==m)ans++;}printf("%d\n",ans);}return 0;}