POJ1655——Balancing Act

Balancing Act

Description

Consider a tree T with N (1 <= N <= 20,000) nodes numbered 1...N. Deleting any node from the tree yields a forest: a collection of one or more trees. Define the balance of a node to be the size of the largest tree in the forest
T created by deleting that node from T.

For example, consider the tree:

Deleting node 4 yields two trees whose member nodes are {5} and {1,2,3,6,7}. The larger of these two trees has five nodes, thus the balance of node 4 is five. Deleting node 1 yields a forest of three trees of equal size: {2,6}, {3,7}, and {4,5}. Each of these
trees has two nodes, so the balance of node 1 is two.

For each input tree, calculate the node that has the minimum balance. If multiple nodes have equal balance, output the one with the lowest number.

Input

The first line of input contains a single integer t (1 <= t <= 20), the number of test cases. The first line of each test case contains an integer N (1 <= N <= 20,000), the number of congruence. The next N-1 lines each contains
two space-separated node numbers that are the endpoints of an edge in the tree. No edge will be listed twice, and all edges will be listed.

Output

For each test case, print a line containing two integers, the number of the node with minimum balance and the balance of that node.

Sample Input

1
7
2 6
1 2
1 4
4 5
3 7
3 1

Sample Output

1 2

Source

POJ Monthly--2004.05.15 IOI 2003 sample task

求树的重心，树的重心定义为，删掉重心以后，树被分为几个部分，使得那几个部分里点最多的那个部分的点数最少

 /*
    定义dp[i]为去掉i结点，剩下的树里，结点最多的那颗树的结点数。
    可分为2类情况。
    1、由于i结点的儿子结点都成了一棵树的根节点，所以dp[i] = (i的每个儿子所拥有的结点数，的最大值)。
    2、而另一种情况就是剩下的那棵树，所以dp[i] = N-num[i]。
    其中num[i]表示以i为根的树的所有结点数，可以dfs求出。
*/
#include <map>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 20010;
const int inf = 0x3f3f3f3f;
int n;

struct node
{
	int next;
	int to;
}edge[N << 1];

int head[N];
int dp[N];
int num[N];
int tot;
bool vis[N];
int dist[N];

void addedge(int from, int to)
{
	edge[tot].to = to;
	edge[tot].next = head[from];
	head[from] = tot++;
}

int dfs(int u)
{
	vis[u] = 1;
	num[u] = 1;
	for (int i = head[u]; ~i; i = edge[i].next)
	{
		int v = edge[i].to;
		if (!vis[v])
		{
			num[u] += dfs(v);
		}
	}
	return num[u];
}

void DP(int u)
{
	vis[u] = 1;
	for (int i = head[u]; ~i; i = edge[i].next)
	{
		int v = edge[i].to;
		if (vis[v])
		{
			dp[u] = max(dp[u], n - num[u]);
		}
		else
		{
			dp[u] = max(dp[u], num[v]);
			DP(v);
		}
	}
}

int main()
{
	int t;
	int u, v;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d", &n);
		memset ( head, -1, sizeof(head) );
		memset ( num, 0, sizeof(num) );
		memset ( vis, 0, sizeof(vis) );
		memset ( dp, 0, sizeof(dp) );
		tot = 0;
		for (int i = 0; i < n - 1; ++i)
		{
			scanf("%d%d", &u, &v);
			addedge(u, v);
			addedge(v, u);
		}
		dfs(1);
		memset ( vis, 0, sizeof(vis) );
		DP(1);
		int ans = inf, id;
		for (int i = 1; i <= n; ++i)
		{
			if (ans > dp[i])
			{
				ans = dp[i];
				id = i;
			}
		}
		printf("%d %d\n", id, ans);
	}
	return 0;
}