POJ 1655：Balancing Act

POJ 1655：Balancing Act

 

Balancing Act Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10311   Accepted: 4261

 

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

表示頭一次接觸樹形dp，一開始的思路想到了要用dfs，並且使用max_i數組表示當前狀況下該節點其孩子節點中最大的值，然後sum數組表示當前狀況下該節點所帶的節點數量，但是這樣的話，就得從每一個節點都要深度搜索一遍，才能得到正確結果，結果提交果然TLE。後來發現深度搜索節點度數為1的就夠了，結果還是TLE。最後，看到別人的代碼，跟我一樣的思路，也是sum數組，還有一個son數組，但是用sum[1]-sum[i]表示了除了當前節點的孩子節點，其父親節點那一個分支段內的節點數量。對這個思路啧啧稱奇，責怪自己有沒有想到，後面的事情就很簡單，之前已經比較過自己的孩子哪一個節點數最多了，這次直接兩兩比較即可。

 

 

代碼:

 

#include   
#include   
#include   
#include   
using namespace std;  

vector  node[20005];  
int result,result_num;  
int used[20005];  
int sum[20005];
int max_i[20005];

int dfs(int i)  
{  
	used[i]=1;  

	int k;  
	sum[i]=0;
	max_i[i]=0;

	for(k=0;kmax_i[i])
			{
				max_i[i]=temp;
			}
		}  
	}
	used[i]=0;	
	return sum[i]+1;
}  
int main()  
{  
	int count,N;
	cin>>count;

	while(count--)
	{
		cin>>N;
		int i,flag;
		result=20005;

		memset(node,0,sizeof(node));
		memset(used,0,sizeof(used));
		for(i=1;i<=N-1;i++)
		{
			int node1,node2;
			cin>>node1>>node2;

			node[node1].push_back(node2);
			node[node2].push_back(node1);
		}
		
		dfs(1);

		for(i=1;i<=N;i++)
		{
			if(max(max_i[i],sum[1]-sum[i])