【POJ 1655】Balancing Act 【树的重心】

标签：poj   树的重心   oi   acm   

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

Output

Sample Input

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

Sample Output

1 2

Source

求树的重心。

任意选一个点为root，进行dfs，找到最大儿子最小的那个结点即可。

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
int tot=0,ans1,n,ans2,h[20005],T,son[20005];
struct edge
{
	int y,ne;
}a[100005];
void Addedge(int x,int y)   //前向星也挺好写的，以后不用vector了
{
	tot++;
	a[tot].y=y;
	a[tot].ne=h[x];
	h[x]=tot;
}
void Dfs(int x)
{
	int i=h[x],now=0;
	son[x]=0;
	while (i)
	{
		int y=a[i].y;
		if (son[y]==-1) 
		{
			Dfs(y);
	 	    son[x]=son[x]+son[y]+1;
		    if (son[y]+1>now) now=son[y]+1;
		}
		i=a[i].ne;
	}
	if (now<n-1-son[x]) now=n-1-son[x];
	if (now==ans1&&ans2>x) ans2=x;
	if (now<ans1)
		ans1=now,ans2=x;
}
int main()
{
    scanf("%d",&T);
	while (T--)
	{
		scanf("%d",&n);
		tot=0;
		for (int i=1;i<=n;i++)
			h[i]=0;
		for (int i=1;i<n;i++)
		{
			int x,y;
			scanf("%d%d",&x,&y);
			Addedge(x,y);
			Addedge(y,x);
		}
		for (int i=1;i<=n;i++)
			son[i]=-1;
		ans1=ans2=n;
		Dfs(1);
		printf("%d %d\n",ans2,ans1);
	}
	return 0;
}

【POJ 1655】Balancing Act 【树的重心】

标签：poj   树的重心   oi   acm   

原文地址：http://blog.csdn.net/regina8023/article/details/41961583