多校Work

Work
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0
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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include<map>
using namespace std;
int t;
int n,k;
int a,b;
int leader[105][105];
int lea[105];
int sum;
void dg(int i)
{
    if(i>=n||lea[i]==0)
        return ;
    for(int j=1; j<=n; j++)
    {
        if(leader[i][j])
        {
            sum+=lea[j];
            dg(j);
        }
    }
}
int main()
{
    int m;
    while(~scanf("%d%d",&n,&k))
    {
        t=0;
        memset(leader,0,sizeof(leader));
        memset(lea,0,sizeof(lea));
        m=n-1;
        while(m--)
        {
            scanf("%d%d",&a,&b);
            leader[a][b]=1;
            lea[a]++;
        }
        for(int i=1; i<=n; i++)
        {
            sum=lea[i];
            dg(i);
            if(sum==k)
                t++;
        }
        printf("%d\n",t);
    }

}