[POJ1655]Balancing Act

题目描述 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. 

输入描述 Input Description

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.

输出描述 Output Description

For each test case, print a line containing two integers, the number of the node with minimum balance and the balance of that node.

样例输入 Sample Input

1
7
2 6
1 2
1 4
4 5
3 7
3 1

样例输出 Sample Output

1 2

数据范围及提示 Data Size & Hint

之前的一些废话：会考结束，没出成绩，准备期末复习。做这种题就是在浪费生命，不如怒刷龙珠。

题解：求树的重心。

代码：

#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<queue>
#include<cstdio>
using namespace std;
typedef long long LL;
#define mem(a,b) memset(a,b,sizeof(a))
typedef pair<int,int> PII;
inline int read()
{
    int x=0,f=1;char c=getchar();
    while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}
    while(isdigit(c)){x=x*10+c-'0';c=getchar();}
    return x*f;
}
const int maxn=20010;
struct Edge
{
    int u,v,next;
    Edge() {}
    Edge(int _1,int _2,int _3):u(_1),v(_2),next(_3) {}
}e[maxn<<1];
int T,n,ce,a,b,first[maxn],size[maxn],A[maxn],ms[maxn],ans;
void addEdge(int a,int b)
{
    e[++ce]=Edge(a,b,first[a]);first[a]=ce;
    e[++ce]=Edge(b,a,first[b]);first[b]=ce;
}
void dfs(int now,int fa)
{
    size[now]=1;
    for(int i=first[now];i!=-1;i=e[i].next)
        if(e[i].v!=fa)
        {
            dfs(e[i].v,now);
            size[now]+=size[e[i].v];
            if(size[e[i].v]>size[ms[now]])ms[now]=e[i].v;
        }
    A[now]=max(size[ms[now]],n-size[now]);
    ans=min(ans,A[now]); 
}
int main()
{
    T=read();
    while(T--)
    {
        mem(first,-1);mem(e,0);mem(A,0);mem(size,0);mem(ms,0);ans=maxn;ce=-1;
        n=read();
        for(int i=1;i<n;i++)a=read(),b=read(),addEdge(a,b);
        dfs(1,0);
        for(int i=1;i<=n;i++)if(A[i]==ans)
        {
            printf("%d %d\n",i,ans);
            break;
        }
    }
}

总结：