poj 1655 Balancing Actツリーの重心を探す

http://poj.org/problem?
id=1655
Balancing Act
Time Limit: 1000MS
 
Memory Limit: 65536K
Total Submissions: 9072
 
Accepted: 3765
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
木の重心を求めます.!
木の重心特性は,この点を根とする有根樹の最大子木の結点数が最も少なく,すなわちこの木をより「バランス」させることであり,easyはこのようにすべての子木の大きさが木全体の大きさの半分を超えないことを知っているので,木の分治時に木が鎖に退化することを防止するのに非常に実用的である.
.
木の重心を求める方法はdfsの簡単な木dpでよい.son[i]がiを根とする子木の結点数(iを含めて葉の結点が1)を表すと、son[i]は+=sum(son[j])…そしてiを根とする最大結点数がmax(son[j],n-son[i]),n-son[i]がiの「上方結点」の個数であることを求める.
木の重心のいくつかの性質について:http://fanhq666.blog.163.com/blog/static/81943426201172472943638/
コード:

/**
 * @author neko01
 */
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <cstring>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
#define min3(a,b,c) min(a,min(b,c))
#define max3(a,b,c) max(a,max(b,c))
#define pb push_back
#define mp(a,b) make_pair(a,b)
#define clr(a) memset(a,0,sizeof a)
#define clr1(a) memset(a,-1,sizeof a)
#define dbg(a) printf("%d
",a)
typedef pair<int,int> pp;
const double eps=1e-9;
const double pi=acos(-1.0);
const int INF=0x3f3f3f3f;
const LL inf=(((LL)1)<<61)+5;
const int N=20005;
struct node{
    int to,next;
}e[N*2];
int head[N];
int son[N];  //son[i]   i           i
int dp[N];   //dp[i]   i           
int tot;
int n;
void init()
{
    tot=0;
    clr1(head);
}
void add(int u,int v)
{
    e[tot].to=v;
    e[tot].next=head[u];
    head[u]=tot++;
}
void dfs(int u,int pre)
{
    son[u]=1;
    dp[u]=0;
    for(int i=head[u];i!=-1;i=e[i].next)
    {
        int v=e[i].to;
        if(v!=pre)
        {
            dfs(v,u);
            son[u]+=son[v];
            dp[u]=max(dp[u],son[v]);
        }
    }
    dp[u]=max(dp[u],n-son[u]);
}
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);
            add(u,v);
            add(v,u);
        }
        dfs(1,0);
        int ans1=0,ans2=n+1;
        for(int i=1;i<=n;i++)
            if(dp[i]<ans2)
            {
                ans2=dp[i];
                ans1=i;
            }
        printf("%d %d
",ans1,ans2);
    }
    return 0;
}

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