建水井


狗家 建水井
https://www.1point3acres.com/bbs/thread-505957-1-1.html

第一轮建水井,一个村庄里有很多房子,有的房子旁边能建水井,有的房子只能通过与其他房子建立管道获得供水。给出了在不同房子旁边建水井的花费,已及两个房子间建管道的花费,要求结果给出一份竞价,使得总花费尽可能少 第一题就卡住了,给出了个不算特别好的算法,我又试了别的想法,但是没想出来,面试官问是要个提示还是就之前我给出的完整的算法写,我选择要了提示,面试官根据我第二个思路引导我构成联通图 最后怕时间不够写代码,提醒说用最小生成树-b

这题一周内已经看到两三次了。仔细想了一下,觉得比最小生成树更复杂,因为可以仅仅找多个联通分量,每个联通分量内只需要一个水井,最后代价最小就行,并不是一定要连通图。

感觉不要去考虑连通分量连通图之类的,直接用最小生成树的思路比较好。因为能建水井的房子,相当于有一条管道可以通往水源,管道长度就是建井费用,最后就是要用最小生成树把所有房子+水源都连接起来。

+1 所以事实上就是构建这样一个graph G=<V, E>, V包括了所有的房子+一个虚拟的水源节点,E包括了所有的管道和打井消耗,然后寻找G的MST即可


考虑到了呀。不要去想水井,就想整个村子只有一个水源,然后只有几个房子可以连到水源。最终要所有房子都有水。

你说的这个情况,那个最后的房子会直接连到水源。但无论如何,都是用一颗树把所有房子还有水源都连起来。

写了一个最小生成树的代码,核心思想就是维护一个PriorityQueue,每次poll出来[cost, v1, v2] 其中v1是已访问的节点,如果v2没有访问就加入到已访问节点中,然后把v2的相连的节点加到queue中。时间复杂度:用邻接表 O( VElogE), 用邻接矩阵是O(VVlogE) 


这是Lazy implementation版的Prim's algorithm
所以是O(E log E)

先建一个dummy,自己家挖水井,当成是这个房子到dummy的一个边,cost就是这个房子自己挖水井的cost,这个时候整个就是一个全连接的图,每个点和其他点都有连接,都有cost,把所有的边加到minheap中,pop cost最小的边,用union find check当前的边可以保留吗?我们的目标是pop出的边能包括所有的房子和dummy,并且没有边(使得房子有重复,会增加cost不是我们想要的) ,每pop一个边,check是不是已经find了?是,不要,不是,留下,一直到这些边包括所有的房子和dummy

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