【HDU】5326 - Work（拓扑）

点击打开题目

Work

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1705    Accepted Submission(s): 1023

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 21 21 32 42 53 63 7

 

Sample Output

2

 

Author

ZSTU

 

Source

2015 Multi-University Training Contest 3

 

用拓扑排序的方法从下层往上层找。

代码如下：

#include <cstdio>#include <cstring>#define CLR(a) memset(a,0,sizeof(a))int main(){int n,k;int sum[111];int in[111];bool re[111][111];int ach;while (~scanf ("%d %d",&n,&k)){CLR(in);CLR(sum);CLR(re);ach = 0;for (int i = 1 ; i < n ; i++){int x,y;scanf ("%d %d",&x,&y);re[y][x] = true;in[x]++;}while (1){for (int i = 1 ; i <= n ; i++){if (in[i] == 0){for (int j = 1 ; j <= n ; j++)if (re[i][j]){re[i][j] = false;in[j]--;sum[j] += sum[i] + 1;break;}in[i] = -1;ach++;}}if (ach == n)break;}int ans = 0;for (int i = 1 ; i <= n ; i++){if (sum[i] == k)ans++;}printf ("%d\n",ans);}return 0;}