poj1655 Balancing Act 找树的重心

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

求树的重心。！

树的重心性质是以该点为根的有根树最大子树的结点数最少，也就是让这树更加"平衡"，easy知道这样全部的子树的大小都不会超过整个树大小的一半，所以在树分治时防止树退化成链非常实用。

。

求树的重心做法是dfs简单的树dp即可。设son[i]表示以i为根的子树的结点数(包含i,叶子结点就是1),那么son[i]就+=sum(son[j])...然后求以i为根的最大结点数就是max(son[j],n-son[i])，n-son[i]是i的"上方结点"的个数。

关于树的重心一些性质:http://fanhq666.blog.163.com/blog/static/81943426201172472943638/

代码：

/**
 * @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]表示以i为根的子树节点个数包含i
int dp[N];   //dp[i]表示以i为根时的最大子树节点数
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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