【poj1655】Balancing Act 求树的重心

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

172 61 21 44 53 73 1

Sample Output

1 2

Source

POJ Monthly–2004.05.15 IOI 2003 sample task

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以前傻X一般的找直径，现在才知道一遍dfs即可。

树的重心的定义：树上找一个点，使它到树上其他点的最大距离最小。或者是以它为根的最大子树最小……总之性质很多就是了。

所以我们可以利用【最大子树最小】的性质，利用dfs找到重心。详见代码。

代码：

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>using namespace std;const int size=100010;const int INF=1<<25;int head[size],nxt[size],to[size],tot=0;void build(int f,int t){    to[++tot]=t;    nxt[tot]=head[f];    head[f]=tot;}int n,x,mindist=INF;int son[size];bool vis[size];void dfs(int u){    son[u]=1;    vis[u]=1;    int sz=0;    for(int i=head[u];i;i=nxt[i])    {        int v=to[i];        if(vis[v]) continue;        dfs(v);        son[u]+=son[v];        sz=max(sz,son[v]);    }    sz=max(sz,n-son[u]);    if(mindist>sz || mindist==sz && u<x )        x=u,mindist=sz;}void init(){    x=0,mindist=INF;    memset(head,0,sizeof(head));    memset(to,0,sizeof(to));    memset(nxt,0,sizeof(nxt));    memset(son,0,sizeof(son));    memset(vis,0,sizeof(vis));}int main(){    int T;    scanf("%d",&T);    while(T--)    {        init();        scanf("%d",&n);        for(int i=1;i<=n-1;i++)        {            int a,b;            scanf("%d%d",&a,&b);            build(a,b); build(b,a);        }        dfs(1);        printf("%d %d\n",x,mindist);    }    return 0;}