hdu 5326 Work（杭电多校赛第三场）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5326

Work

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 583    Accepted Submission(s): 392

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 

 

题目大意：输入的第一行：第一个数字n表示的是有几个人，第二个数字m表示查找管理两个人的。

　　   　　接下去就有n-1行表示前面的管理后面的，最后求的就是有几个人是管理m个人的。

 

详见代码。

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int n,m,Map[110][110];

int dfs(int x)
{
    int sum=0;
    for (int i=1;i<=n;i++)
    {
        if (Map[x][i]==1)
        {
            sum++;
            sum+=dfs(i);
        }
    }
    return sum;
}

int main()
{
    int a,b,ans;
    while (~scanf("%d%d",&n,&m))
    {
        ans=0;
        memset(Map,0,sizeof(Map));
        //memset(ok,0,sizeof(ok));
        for (int i=1; i<=n-1; i++)
        {
            scanf("%d%d",&a,&b);
            Map[a][b]=1;
        }
        for (int i=1;i<=n;i++)
        {
            if (dfs(i)==m)
                ans++;
        }
        printf ("%d\n",ans);
    }
    return 0;
}