poj 1655 树形dp求取树的重心

http://poj.org/problem?id=1655

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

/**
poj 1655  利用树形dp求树的重心
题目大意：求数的重心
解题思路：树的重心定义为:找到一个点,其所有的子树中最大的子树节点数最少,那么这个点就是这棵树的重心,删去重
          心后，生成的多棵树尽可能平衡.  实际上树的重心在树的点分治中有重要的作用, 可以避免N^2的极端复杂度（从退化链的一端出发），保证
          NlogN的复杂度, 利用树形dp可以很好地求树的重心.
*/
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn=20005;

struct note
{
    int v,next;
}edge[maxn*2];

int head[maxn],ip;
int maxx,point,n,num[maxn];

void init()
{
    memset(head,-1,sizeof(head));
    ip=0;
}

void addedge(int u,int v)
{
    edge[ip].v=v,edge[ip].next=head[u],head[u]=ip++;
}

void dfs(int u,int pre)
{
    int tmp=-0x3f3f3f3f;
    num[u]=1;
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(v==pre)continue;
        dfs(v,u);
        num[u]+=num[v];
        tmp=max(tmp,num[v]);
    }
    tmp=max(tmp,n-num[u]);///除去以u为根节点的子树部分剩下的节点数
    if(tmp<maxx||tmp==maxx&&u<point)
    {
        point=u;
        maxx=tmp;
    }
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        init();
        for(int i=1;i<n;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            addedge(u,v);
            addedge(v,u);
        }
        memset(num,0,sizeof(num));
        maxx=0x3f3f3f3f;
        dfs(1,-1);
        printf("%d %d\n",point,maxx);
    }
    return 0;
}