Google – Average Length of City Path



Google – Average Length of City Path

给n个城市,和n – 1条edge, 这些edges可以保证所有城市都两两互通,每条edge都有长度,求所有城市两两之间path的长度之和,再除以path的数量得到的平均数。

[Solution]
第一种解法就是brute force,O(n^2) time
第二种解法来自一亩三分地,可以在O(n)时间内解决问题。思路非常tricky,类似于Leetcode 310 Minimum Height Tree的解法。

因为题目明确给出了有n个node和n – 1条edges,看到这两个数字应该很本能的反应出这道题和tree拖不了关系。但是Brute Force的解法并没有利用到这个条件。而O(n)的解法正是利用构建Minimum Height Tree来解决问题。下面是算法思路:

(1)算degree
(2)逐层删leaf node来找root。就是Minimum Height Tree的算法。
(3)在删leaf node的过程中多做两件事:建tree和计算sub tree size(包括该节点本身)。如果在第(2)步有两个root,随便选一个作为root,另一个作为子节点就好。这里的主要目的是建Tree.
(4)如果总共有N个结点,当前结点v的sub tree size为K (包括v本身),那么连接v和其parent的那条edge就会被遍历K * (N-K)次。
这是因为如果以v为分界把这个graph分为两部分,一部分有K个点,另一部分为N-K个点。因为每对Pair都需要走一遍,也就是左边的K个点中的任何一个都要去找右边的N-K个点。又因为树的特性,每对pair之间只可能有一条path,所以v到其parent之间的这条edge必定会被遍历K * (N-K)次。
(5)构建完tree,证明完(4),剩下的只需要对tree做一个dfs traversal,就可以得到总的path sum。再除以n-1就可以得到答案了。


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