解题报告 之 HDU5326 Work

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 

题目大意：一个公司，每个人（除了老板）都有一个直接上司，上司的上司还是你的间接上司。给你一个关系树的每条边，问你树中有多少个节点的子孙数为k？（直接下属+间接下属）。

分析：可以看到数据规模很小，那么就是一道深度搜索的水题，可以算是图论很大众的知识了。结点的下属数 = 直接下属数 + 每个下属的下属数。那么就是一个递归了。

上代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<cstring>
using namespace std;

const int MAXN = 1e2+10;

vector<int> node[MAXN];
int deg[MAXN];
int ma[MAXN];

int dfs(int u)
{
	if(node[u].size() == 0)
	{
		ma[u] = 0;
		return ma[u] + 1;
	}

	for(int i = 0; i < node[u].size(); i++)
		ma[u] += dfs( node[u][i] );

	return ma[u] + 1;
}

int main()
{
	int n, k;
	while(scanf( "%d%d", &n, &k ) == 2)
	{
		for(int i = 0; i < MAXN; i++)
			node[i].clear();
		memset( deg, 0, sizeof deg );
		memset( ma, 0, sizeof ma );

		int a, b,root;
		for(int i = 0; i <n-1; i++)
		{
			scanf( "%d%d", &a, &b );
			node[a].push_back( b );
			deg[b]++;
		}

		for(int i = 1; i <= n; i++)
		{
			if(deg[i] == 0)
			{
				root = i;
				break;
			}
		}
		dfs( 1 );
		int ans = 0;
		for(int i = 1; i <= n; i++)
		{
			if(ma[i] == k)
				ans++;
		}
		printf( "%d\n", ans );
	}
	return 0;
}

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