?
?
Balancing Act
| Time Limit: 1000MS |
? |
Memory Limit: 65536K |
| Total Submissions: 9072 |
? |
Accepted: 3765 |
?
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
求树的重心！！
树的重心性质是以该点为根的有根树最大子树的结点数最少，也就是让这树更加平衡，容易知道这样所有的子树的大小都不会超过整个树大小的一半，所以在树分治时防止树退化成链很有用。。
求树的重心做法是dfs简单的树dp就行。设son[i]表示以i为根的子树的结点数(包括i,叶子结点就是1),那么son[i]就+=sum(son[j])...然后求以i为根的最大结点数就是max(son[j],n-son[i])，n-son[i]是i的上方结点的个数。
关于树的重心一些性质:http://fanhq666.blog.163.com/blog/static/81943426201172472943638/
代码：
?
/**
* @author neko01
*/
//#pragma comment(linker, /STACK:102400000,102400000)
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include