【HDU】5326 - Work（拓扑）

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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 2
1 2
1 3
2 4
2 5
3 6
3 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;
}