如何从“不相交集"中获取所有元素的列表. [英] How to get list of all elements from a 'Disjoint Sets'

问题描述

在我的问题中，我有一堆元素(类元素).假设我有1000个元素.这些元素最初是未关联的，这意味着它们在自己的集合中.

In my problem, I have a bunch of Elements (Class Element). Say I have 1000 Elements. These Elements are initially un-associated which means they are in their own sets.

稍后，我需要使用并集操作根据我的程序流来合并其中的一些集合.我计划使用Boost库的disjoint_set( http://www.boost.org/doc/libs/1_57_0/libs/disjoint_sets/disjoint_sets.html )

Later I need to use the union operation to merge some of these sets based of my program flow. I plan to use the boost library's disjoint_set (http://www.boost.org/doc/libs/1_57_0/libs/disjoint_sets/disjoint_sets.html)

我的问题是，在给定集合代表的情况下，如何列出集合中的元素.

My question is how is it possible to list the Elements in a set given the representative of the set.

为此类任务设置了最佳的数据结构.那我应该考虑使用其他东西吗?

Is disjoint_set the best data structure for such a task. So should I look into using something else?

推荐答案

从您的散文描述中，我不知道这些集合实际上会形成任何图形.

From your prose description, I get no information that the sets would actually form any graphs.

如果您要做的只是将元素与集合相关联，我建议

If all you want to do is associate Elements with a set, I'd suggest

std::map<ElementId, SetId>

(如果您知道指针仍然有效，则 ElementId 可以简单地是 Element * ).

(where ElementId could simply be Element* if you know the pointers stay valid).

如果您还希望能够高效地查询逆数

If you also want to be able to query the inverse efficiently

bimap<Element, bimaps::multiset_of<SetId> >

将是候选人.观看演示 Coliru现场直播 ¹

would be a candidate. See a demonstration Live On Coliru¹

#include <boost/range/iterator_range.hpp>
#include <boost/bimap/multiset_of.hpp>
#include <boost/bimap.hpp>
#include <iostream>

using namespace boost;

int main() {
    using Element = int; // for simplicity :)
    using SetId   = int;
    using Sets = bimap<Element, bimaps::multiset_of<SetId> >;

    Sets sets;
    sets.insert({ Element(1), 300 });
    sets.insert({ Element(2), 300 });
    sets.insert({ Element(3), 400 });
    sets.insert({ Element(4), 300 });

    // give us set #300
    for (auto& e : make_iterator_range(sets.right.equal_range(300)))
        std::cout << e.first << " - Element(" << e.second << ")\n";
}

打印

300 - Element(1)
300 - Element(2)
300 - Element(4)

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¹Coliru似乎情绪低落.以后会添加

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¹ Coliru seems down. Will add later

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