poj 1655 Balancing Act（树dp）

Balancing Act

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8095		Accepted: 3341

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

题意：去掉一个点，使其得到的子树的最大结点数最小。。。换言之就是求树的重心

题解：树dp，用dfs求一下son，然后对于每个结点分出的最大子树节点数有res=MAX（res，son【to】）和res=MAX（res，n-son【x】）

#include<stdio.h>
#include<string.h>
#define MAXN 20008
struct edge{
    int to,next;
}p[MAXN*2];
int head[MAXN],son[MAXN],all,n,res1,res2;
int MAX(int x,int y){ return x>y?x:y; }
int MIN(int x,int y){ return x<y?x:y; }
void add(int x,int y)
{
    p[all].to=y;
    p[all].next=head[x];
    head[x]=all++;
}
void dfs(int x,int f)
{
    int res=0,i;

    for(i=head[x];i!=-1;i=p[i].next)
    {
        if(p[i].to==f) continue;
        dfs(p[i].to,x);
        res=MAX(res,son[p[i].to]);
        son[x]+=son[p[i].to];
    }
    son[x]++;
    res=MAX(res,n-son[x]);
    if(res<res2||(res==res2&&x<res1)) res1=x,res2=res;
}
int main()
{
    int t,i,x,y;

    scanf("%d",&t);
    while(t--)
    {
        memset(head,-1,sizeof(head));
        memset(son,0,sizeof(son));
        scanf("%d",&n);
        for(all=0,i=1;i<n;i++)
        {
            scanf("%d%d",&x,&y);
            add(x,y);  add(y,x);
        }
        res1=res2=1000000;
        dfs(1,-1);
        printf("%d %d\n",res1,res2);
    }

    return 0;
}