# hdoj 5326 Work （简单并查集） 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.

InputThere 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
OutputFor 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

#include<iostream>

using namespace std;
int n,k;
int s;
int par;
void init(){
for(int i=0;i<101;i++){
par[i]=i;
s[i]=0;
}
}
void setp(int a,int b){
par[b]=a;
s[a]+=s[b]+1;
int t=a;
while(par[t]!=t){
t=par[t];
s[t]+=s[b]+1;
}

}
int main(){
int a,b;
while(cin>>n>>k){
init();
int sum=0;
for(int i=1;i<n;i++){
cin>>a>>b;
setp(a,b);
}
for(int i=1;i<=n;i++){
if(s[i]==k){
sum++;
}
}
cout<<sum<<endl;
}

return 0;
}