poj1655 Balancing Act 求树的重心

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

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！！

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"，容易知道这样所有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很有用。。

求树的重心做法是dfs简单的树dp就行。设son[i]表示以i为根的子树的结点数(包括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\n",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括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\n",ans1,ans2);
    }
    return 0;
}

poj1655 Balancing Act 求树的重心

原文：http://blog.csdn.net/neko01/article/details/39940939