如何实现由链表不相交集的数据结构？

本文介绍了如何实现由链表不相交集的数据结构？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我不知道如何通过链表来Implemente不相交集数据结构？谁可以告诉我吗？ 我特别感到困惑如何查找集（）可以工作O（1）时间，这样的实现。THX〜

i don't know how to Implemente Disjoint-set data structure by Linked list？ who can tell me？ i'm especially confused about how can the find-set（） can work O(1) time in such implementation。 thx~

推荐答案

虽然不相交集数据结构通常被解释为对本之木，它通常实施使用数组：一个数组的索引映射到指数的父母和一个阵列映射的索引，它的等级。此摆脱了很多无谓的开销（在空间和时间），这是不相关的理论，但它是在实际中。您可能需要一种方法来某种对象的索引和实例之间的映射。

While the disjoint-set data structure is typically explained as a forest of trees, it's typically implemented with arrays: one array to map an index to the index of its parent, and one array to map an index to its rank. This gets rid of a lot of pointless overhead (in both space and time), which isn't relevant in theory, but it is in practice. You may need a way to map between the indexes and instances of some kind of object.

查找不为O（1），由工作方式。跟了上来prevents诀窍查找采取线性时间，但它仍然可以在最坏的情况下步骤的对数数。在O（α（n））的时间是摊销的时间，它的棘手解释它从何而来 - 你可以找到分析的效率不错，但不是线性并集算法[的Tarjan]

find does not work in O(1), by the way. The trick with the ranks prevents find from taking linear time, but it can still take a logarithmic number of steps in the worst case. The O(α(n)) time is an amortized time, and it's tricky to explain where it comes from - you can find the analysis in Efficiency of a Good But Not Linear Set Union Algorithm [Tarjan].

下面是它可以工作的一种方式：

Here's one way it could work:

disjoint_set { int[] parent, rank; makeset(int n) { rank = new int[n]; parent = new int[n]; for(int i = 0; i < n; i++) parent[i] = i; } int find(int i) { if (parent[i] != i) parent[i] = find(parent[i]); return parent[i]; } void union(int x, int y) { x_root = find(x); y_root = find(y); if (x_root != y_root) { if (rank[x_root] < rank[y_root]) parent[x_root] = y_root; else if (rank[x_root] > rank[y_root]) parent[y_root] = x_root; else { parent[y_root] = x_root; rank[x_root]++; } } } }