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：
给你一颗n个节点的数，让你求出树的重心以及以重心为根的子树中最大的子树节点个数。
思路：
紧扣树的重心的定义：
定义为：在树中招到一个节点，其所有子树中最大的子树节点个数最小，那么这个点就是树的重心，删除重心后，生成的多颗树尽可能的平衡。
性质：
1、树中所有点到某个点的距离的sum和，到重心的距离sum和是最小的。如果有两个重心，那么他们的距离sum和是一样的、
2、把两棵树通过一个边连接得到一颗新的树，那么新的树的重心在原来两个树的重心的路径上。
3、把一个树添加或删除一个叶子节点，那么它的重心最多移动一个边的距离。
我们根据“所有子树中最大的子树节点个数最小”，这个性质，在dfs整颗树的时候，用cntson[i] 数组记录第i个节点为根的子树节点个数
用在dfs函数用，用cnt记录 当前节点的最大子树节点个数。
那么直需要维护cnt中的最小值即可得到答案。
细节见代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), ' ', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int* p);
const int maxn = 100010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
std::vector<int> son[maxn];
int t;
int n;
int ans1;
int ans2;
int cntson[maxn];
void dfs(int x,int pre)
{
cntson[x]=1;
int cnt=0;
// for(auto y:son[x])
for(int i=0;i<sz(son[x]);++i)
{
int y=son[x][i];
if(y!=pre)
{
dfs(y,x);
cntson[x]+=cntson[y];
cnt=max(cnt,cntson[y]);
}
}
cnt=max(cnt,n-cntson[x]);
if(cnt<ans2||(cnt==ans2&&x<ans1))
{
ans2=cnt;
ans1=x;
}
}
void init()
{
repd(i,1,n)
{
son[i].clear();
}
ans2=inf;
}
int main()
{
//freopen("D:\code\text\input.txt","r",stdin);
//freopen("D:\code\text\output.txt","w",stdout);
scanf("%d",&t);
while(t--)
{
init();
scanf("%d",&n);
repd(i,2,n)
{
int u,v;
scanf("%d %d",&u,&v);
son[v].push_back(u);
son[u].push_back(v);
}
dfs(1,1);
printf("%d %d
",ans1,ans2 );
}
return 0;
}
inline void getInt(int* p) {
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '
');
if (ch == '-') {
*p = -(getchar() - '0');
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 - ch + '0';
}
}
else {
*p = ch - '0';
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 + ch - '0';
}
}
}