poj 1655 樹形dp求取樹的重心

poj 1655 樹形dp求取樹的重心

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 whZ喎?http://www.Bkjia.com/kf/ware/vc/" target="_blank" class="keylink">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"pst">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  利用樹形dp求樹的重心
題目大意：求數的重心
解題思路：樹的重心定義為:找到一個點,其所有的子樹中最大的子樹節點數最少,那麼這個點就是這棵樹的重心,刪去重
          心後，生成的多棵樹盡可能平衡.  實際上樹的重心在樹的點分治中有重要的作用, 可以避免N^2的極端復雜度（從退化鏈的一端出發），保證
          NlogN的復雜度, 利用樹形dp可以很好地求樹的重心.
*/
#include 
#include 
#include 
#include 
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為根節點的子樹部分剩下的節點數
    if(tmp