2017ACM/ICPC亚洲区沈阳站 L.Tree （dfs）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6228

Problem Description
Consider a un-rooted tree T which is not the biological significance of tree or plant, but a tree as an undirected graph in graph theory with n nodes, labelled from 1 to n. If you cannot understand the concept of a tree here, please omit this problem.
Now we decide to colour its nodes with k distinct colours, labelled from 1 to k. Then for each colour i = 1, 2, · · · , k, define Ei as the minimum subset of edges connecting all nodes coloured by i. If there is no node of the tree coloured by a specified colour i, Ei will be empty.
Try to decide a colour scheme to maximize the size of E1 ∩ E2 · · · ∩ Ek, and output its size.

Input
The first line of input contains an integer T (1 ≤ T ≤ 1000), indicating the total number of test cases.
For each case, the first line contains two positive integers n which is the size of the tree and k (k ≤ 500) which is the number of colours. Each of the following n - 1 lines contains two integers x and y describing an edge between them. We are sure that the given graph is a tree.
The summation of n in input is smaller than or equal to 200000.

Output
For each test case, output the maximum size of E1 ∩ E1 … ∩ Ek.

Sample Input

3
4 2
1 2
2 3
3 4
4 2
1 2
1 3
1 4
6 3
1 2
2 3
3 4
3 5
6 2

Sample Output

1
0
1

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题意

给出一棵树，可以给书上的节点染成 k 种不同的颜色，每种颜色的集合为相同颜色所经过的边，求每种颜色集合的交集的最大值。

分析

如果一条边能被算入所有颜色集合的交集，那么它的两侧肯定都分别包含所有颜色。设 sum[x] 表示以 x 为根节点的子树所包含的节点数，那么如果 sum[x] >= k && n - sum[x] >= k ， x 节点到其父节点（设从 1 开始dfs）这条边必被选中。

代码

#include<bits/stdc++.h>
using namespace std;

int t,n,k;
vector<int> g[200007];
int sum[200007];
int ans;

void dfs(int x,int fa)
{
	sum[x] = 1;
	int len = g[x].size();
	for(int i=0;i<len;i++)
	{
		int to = g[x][i];
		if(fa == to) continue;
		dfs(to, x);
		sum[x] += sum[to];
	}
	if(sum[x] >= k && n - sum[x] >= k)
	{
		ans++;
	}
}

int main()
{
	scanf("%d",&t);
	while(t--)
	{
		ans = 0;
		scanf("%d%d",&n,&k);
		for(int i=1;i<=n;i++) g[i].clear();
		for(int i=1;i<n;i++)
		{
			int u,v;
			scanf("%d%d",&u,&v);
			g[u].push_back(v);
			g[v].push_back(u);
		}
		if(n < 2 * k)
		{
			printf("0\n");
		}
		else
		{
			dfs(1, 0);
			printf("%d\n",ans);
		}
	}
	return 0;
}