POJ - 1655 Balancing Act（树的重心）

Time limit：1000 ms；Memory limit：65536 kB

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

题意：n个结点的一棵树，去掉某个结点x后，会变成一个森林。这个森林中的每棵树都有自己的结点数，其中最大结点数设为ans，问删除哪个结点x后，ans最小？并写出x，ans。

分析：其实就是求树的重心（定义：对于一棵n个结点的无根树，找到一个点，使得把树变成以该点为根的有根树时，最大子树的结点数最小。换句话说，删除这个点后最大连通块（一定是树）的结点数最小。）

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解题步骤如下：

①题目给的是无向图，所以用邻接表储存时记得双向连边。

②用一次DFS找到重心，并求出其最大连通块的结点数。

注意事项：任取一点为根时，一定要标记走过的点（记忆化处理，不走回头路）

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AC代码如下（157ms）：

/*
 * dp[i]表示以i为根的子树的结点个数
 * dp[i] = Σdp[j] + 1; j是i的son
 * 结点i的最大连通块结点个数为max(dp[i],N-dp[i])
 */
#include <vector>
#include <algorithm>
#include <stdio.h>
using namespace std;
#define fast                                ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define ll                                  long long
#define _for(i,a,b)                         for(int i = a;i < b;i++)
#define rep(i,a,b)                          for(int i = a;i <= b;i++)

const int maxn = 20010;
int kase, N, ans[maxn];
vector<int>G[maxn];//邻接表
int dp[maxn];
bool flag[maxn];//记忆化处理。

void dfs(int r)
{
	int sum, k;
	sum = k = 0;
	flag[r] = true;//标记经历过此点（此处一定要标记，否则死循环）
	int nson = G[r].size();//此点儿子的个数
	_for(i, 0, nson)
	{
		int son = G[r][i];
		if (!flag[son])
		{
			dfs(son);
			sum += dp[son];
			k = max(k, dp[son]);
		}
		else continue;
	}
	dp[r] = sum + 1;//以r为根的结点总个数
	ans[r] = max(k, N - dp[r]);//以r为根的最大连通块的结点个数
}

int main()
{
	scanf("%d", &kase);
	while (kase--)
	{
		scanf("%d", &N);
		rep(i, 1, N)G[i].clear();//清空邻接表
		memset(dp, 0, sizeof(dp));
		memset(flag, false, sizeof(flag));
		int x, y;
		_for(i, 0, N - 1)
		{
			scanf("%d%d", &x, &y);
			G[x].push_back(y);
			G[y].push_back(x);
		}
		int dot = 0;
		int cnt = maxn;
		dfs(1);
		rep(i, 1, N)
		{
			if (ans[i]<cnt)
				dot = i, cnt = ans[i];
		}

		printf("%d %d\n", dot, cnt);
	}
	return 0;
}