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。

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4  5 using namespace std; 6  7 int n,m,Map[110][110]; 8  9 int dfs(int x)10 {11     int sum=0;12     for (int i=1;i<=n;i++)13     {14         if (Map[x][i]==1)15         {16             sum++;17             sum+=dfs(i);18         }19     }20     return sum;21 }22 23 int main()24 {25     int a,b,ans;26     while (~scanf("%d%d",&n,&m))27     {28         ans=0;29         memset(Map,0,sizeof(Map));30         //memset(ok,0,sizeof(ok));31         for (int i=1; i<=n-1; i++)32         {33             scanf("%d%d",&a,&b);34             Map[a][b]=1;35         }36         for (int i=1;i<=n;i++)37         {38             if (dfs(i)==m)39                 ans++;40         }41         printf ("%d\n",ans);42     }43     return 0;44 }