poj1655Balancing Act（树的重心）

Language: Default Balancing Act Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9418   Accepted: 3941 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

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __int64 ll;

using namespace std;

#define N 20005

int num[N],dp[N];
int head[N],k;
int n;

struct stud{
  int to,next;
}e[N*2];

inline void add(int u,int v)
{
	e[k].to=v;
	e[k].next=head[u];
	head[u]=k++;
}

void dfs(int x,int pre)
{
    dp[x]=0;
    num[x]=1;
    int i;
    for(i=head[x];i!=-1;i=e[i].next)
	{
		int to=e[i].to;
		if(to==pre) continue;
		dfs(to,x);

		dp[x]=max(dp[x],num[to]);
		num[x]+=num[to];
	}

	dp[x]=max(dp[x],n-num[x]); //例如树是 1 2 3 4 5,没有这一步 dp[5]就是0
}

int main()
{
	int i,j,t;
	scanf("%d",&t);
	while(t--)
	{
		k=0;
		memset(head,-1,sizeof(head));

		scanf("%d",&n);

		int u,v;
		for(i=1;i<n;i++)
		{
			scanf("%d%d",&u,&v);
			add(u,v);
			add(v,u);
		}

		dfs(1,-1);

		int ans1=1,ans2=dp[1];

		for(i=2;i<=n;i++)
			if(dp[i]<ans2)
		{
			ans1=i;
			ans2=dp[i];
		}

		printf("%d %d\n",ans1,ans2);

	}
	return 0;
}