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[][];

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

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