POJ1655：Balancing Act(树形DP)

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

题意：建立一棵树，要求删除一个节点后，得到的几棵树节点数平衡，而且节点数要尽可能多，输出删除的节点与节点数

思路：树形DP，记录每个点的节点数与该点子树的最大节点数，进行搜索

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

struct node
{
    int now,next;
} tree[40005];

int n,sum[40005],son[20005];
int head[20005],vis[20005],len;
//sum数组代表包含该节点与该点子节点数量和
//son数组代表该节点包含的所有子树的最大节点数
void add(int x,int y)//建树
{
    len++;
    tree[len].now = y;
    tree[len].next = head[x];
    head[x] = len;

    len++;
    tree[len].now = x;
    tree[len].next = head[y];
    head[y] = len;
}

void dfs(int root)
{
    int i,j;
    vis[root] = 1;
    for(i = head[root]; i!=-1; i = tree[i].next)
    {
        if(!vis[tree[i].now])
        {
            dfs(tree[i].now);
            sum[root]+=sum[tree[i].now];
            if(son[root]<sum[tree[i].now])
                son[root] = sum[tree[i].now];
        }
    }
}


int main()
{
    int T,i,j,x,y,ans,pos;
    scanf("%d",&T);
    while(T--)
    {
        len = 0;
        scanf("%d",&n);
        memset(tree,0,sizeof(tree));
        memset(head,-1,sizeof(head));
        for(i = 1; i<n; i++)
        {
            scanf("%d%d",&x,&y);
            add(x,y);
        }
        memset(son,0,sizeof(son));
        memset(vis,0,sizeof(vis));
        for(i = 0; i<=n; i++)
            sum[i] = 1;
        dfs(1);
        pos = 1;
        ans = son[1];
        for(i = 2; i<n; i++)
        {
            if(ans>max(sum[1]-sum[i],son[i]))
            {
                ans = max(sum[1]-sum[i],son[i]);
                pos = i;
            }
        }
        printf("%d %d\n",pos,ans);
    }

    return 0;
}