2019CCPC-江西省赛 A-Cotree(树的重心)

解决2019年CCPC江西省赛中的A题,涉及树的重心概念。给定两棵不连通的树,目标是添加一条边将它们连通,使得树上所有节点两两间的最短路径长度之和最小。题目要求输出这个最小和。解题策略是找到每棵树的重心,然后计算连接重心后的距离和。

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链接:JXCPC A-Cotree

Avin has two trees which are not connected. He asks you to add an edge between them to make them connected while minimizing the function ∑ i = 1 n − 1 ∑ j = i + 1 n d i s t a n c e ( i , j ) \sum_{i=1}^{n-1}\sum_{j=i+1}^{n}distance(i,j) i=1n1j=i+1ndistance(i,j), where dis(i, j) represents the number of edges of the path from i to j. He is happy with only the function value.

Input

The first line contains a number n (2<=n<=100000). In each of the following n−2 lines, there are two numbers u and v, meaning that there is an edge between u and v. The input is guaranteed to contain exactly two trees.

Output

Just print the minimum function value.

Sample Input

3
1 2

Sample Output

4



题意:

给出两棵树(共n个结点,即n-2条边),加一条边连接两棵树使得得到的新树上结点两两距离和最小。即 ∑ i = 1 n − 1 ∑ j = i + 1 n d i s t a n c e ( i , j ) \sum_{i=1}^{n-1}\sum_{j=i+1}^{n}distance(i,j) i=1n1

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