多校3 1011 Work

Work

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 230    Accepted Submission(s): 171

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

 

 

Source

2015 Multi-University Training Contest 3

 

 

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#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int n,k,s;
int a[105][105];

int dfs(int g)
{
    for(int i=1;i<=n;i++)
    {
        if(a[g][i]==1)
        {
            s++;
            dfs(i);
        }
    }
    return 0;
}

int main()
{
    int i,j,x,y;
    while(scanf("%d %d",&n,&k)!=EOF)
    {
        memset(a,0,sizeof(a));
        for(i=1;i<=n-1;i++)
        {
            scanf("%d %d",&x,&y);
            a[x][y]=1;
        }
        int ans=0;
        for(i=1;i<=n;i++)
        {
            s=0;
            dfs(i);
            if(s==k)
                ans++;
        }
        printf("%d\n",ans);
    }
}