poj 1655 树形dp

题意相当于给你一棵树   让你求每个点的不同子树上的节点个数吧   注意存边的时候存双向边

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

struct node
{
	int to,next;
}A[40010];
int tot,list[20010],mark[20010],n;
int add(int a,int b)
{
	A[++tot].to = b;
   	A[tot].next = list[a];
   	list[a] = tot;
 	return 0;
}
int dp[20010],subtree[20010];
int max(int a,int b)
{
	return a>b?a:b;
}
int dfs(int point)
{
	mark[point]=1;
	int b;
	int k=list[point];
	if(k==-1) {dp[point]=n-1;subtree[point]=0;return 0;}
	for(;k!=-1;k = A[k].next)
   	{
	   	b = A[k].to;
	   	if(mark[b]) continue;
	   	dfs(b);
	   	subtree[point]+=subtree[b]+1;
	   	dp[point]=max(dp[point],subtree[b]+1);
   	}
   	dp[point]=max(dp[point],n-subtree[point]-1);
   	return 0;

}
int main()
{
	int i,j,T;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&n);
		tot=1;
		memset(list,-1,sizeof(list));
		memset(mark,0,sizeof(mark));
		for(i=1;i<n;i++)
		{
			int a,b;
			scanf("%d%d",&a,&b);
			add(a,b);
			add(b,a);
		}
		memset(dp,0,sizeof(dp));
		memset(subtree,0,sizeof(subtree));
		memset(mark,0,sizeof(mark));
		dfs(1);
		int Min=3000000,k;
		for(i=1;i<=n;i++)
		if(dp[i]<Min)
		{
			Min=dp[i];
			k=i;
		}
		printf("%d %d\n",k,Min);
	}
	return 0;
}