在C ++中实现不相交集(Union Find)

时间:2010-12-21 11:38:32

标签: c++ disjoint-sets

我正在尝试实现用于Kruskal算法的Disjoint Sets,但我无法确切地了解它应该如何完成,特别是如何管理树林。在阅读了维基百科关于不相交集的描述之后,在阅读了算法导论(Cormen等人)中的描述后,我得出了以下内容:

    class DisjointSet
    {
    public:
        class   Node 
        {
        public:
            int         data;
            int         rank;

            Node*       parent;

            Node() : data(0), 
                     rank(0),
                     parent(this) { } // the constructor does MakeSet
        };

        Node*   find(Node*);
        Node*   merge(Node*, Node*); // Union
    };


    DisjointSet::Node*   DisjointSet::find(DisjointSet::Node* n)
    {
        if (n != n->parent) {
            n->parent = find(n->parent);
        }
        return n->parent;
    }

    DisjointSet::Node* DisjointSet::merge(DisjointSet::Node* x,
                                          DisjointSet::Node* y)
    {
        x = find(x);
        y = find(y);

        if (x->rank > y->rank) {
            y->parent = x;
        } else {
            x->parent = y;
            if (x->rank == y->rank) {
                ++(y->rank);
            }
        }
    }

我很确定这是不完整的,我错过了一些东西。

算法简介提到应该有一片树林,但它没有给出这片林的实际实施的任何解释。我观看了CS 61B第31讲:不相交集(http://www.youtube.com/watch?v=wSPAjGfDl7Q),这里讲师只使用一个数组来存储森林及其所有树和值。我提到的没有明确的'Node'类型。我还发现了许多其他来源(我不能发布多个链接),这也使用了这种技术。我很乐意这样做,除了这依赖于数组的索引进行查找,因为我想存储除int之外的类型的值,我需要使用别的东西(std :: map可以想到)。 / p>

我不确定的另一个问题是内存分配,因为我使用的是C ++。我存储指针树,我的MakeSet操作将是:new DisjointSet :: Node; 。现在,这些节点只有指向父母的指针,所以我不确定如何找到树的底部。我怎样才能遍历我的树木以解除所有这些?

我理解这个数据结构的基本概念,但我对实现有点困惑。非常欢迎任何意见和建议,谢谢。

10 个答案:

答案 0 :(得分:4)

无论如何都不是一个完美的实现(毕竟我写了它!),但这有帮助吗?

/***
 * millipede: DisjointSetForest.h
 * Copyright Stuart Golodetz, 2009. All rights reserved.
 ***/

#ifndef H_MILLIPEDE_DISJOINTSETFOREST
#define H_MILLIPEDE_DISJOINTSETFOREST

#include <map>

#include <common/exceptions/Exception.h>
#include <common/io/util/OSSWrapper.h>
#include <common/util/NullType.h>

namespace mp {

/**
@brief  A disjoint set forest is a fairly standard data structure used to represent the partition of
        a set of elements into disjoint sets in such a way that common operations such as merging two
        sets together are computationally efficient.

This implementation uses the well-known union-by-rank and path compression optimizations, which together
yield an amortised complexity for key operations of O(a(n)), where a is the (extremely slow-growing)
inverse of the Ackermann function.

The implementation also allows clients to attach arbitrary data to each element, which can be useful for
some algorithms.

@tparam T   The type of data to attach to each element (arbitrary)
*/
template <typename T = NullType>
class DisjointSetForest
{
    //#################### NESTED CLASSES ####################
private:
    struct Element
    {
        T m_value;
        int m_parent;
        int m_rank;

        Element(const T& value, int parent)
        :   m_value(value), m_parent(parent), m_rank(0)
        {}
    };

    //#################### PRIVATE VARIABLES ####################
private:
    mutable std::map<int,Element> m_elements;
    int m_setCount;

    //#################### CONSTRUCTORS ####################
public:
    /**
    @brief  Constructs an empty disjoint set forest.
    */
    DisjointSetForest()
    :   m_setCount(0)
    {}

    /**
    @brief  Constructs a disjoint set forest from an initial set of elements and their associated values.

    @param[in]  initialElements     A map from the initial elements to their associated values
    */
    explicit DisjointSetForest(const std::map<int,T>& initialElements)
    :   m_setCount(0)
    {
        add_elements(initialElements);
    }

    //#################### PUBLIC METHODS ####################
public:
    /**
    @brief  Adds a single element x (and its associated value) to the disjoint set forest.

    @param[in]  x       The index of the element
    @param[in]  value   The value to initially associate with the element
    @pre
        -   x must not already be in the disjoint set forest
    */
    void add_element(int x, const T& value = T())
    {
        m_elements.insert(std::make_pair(x, Element(value, x)));
        ++m_setCount;
    }

    /**
    @brief  Adds multiple elements (and their associated values) to the disjoint set forest.

    @param[in]  elements    A map from the elements to add to their associated values
    @pre
        -   None of the elements to be added must already be in the disjoint set forest
    */
    void add_elements(const std::map<int,T>& elements)
    {
        for(typename std::map<int,T>::const_iterator it=elements.begin(), iend=elements.end(); it!=iend; ++it)
        {
            m_elements.insert(std::make_pair(it->first, Element(it->second, it->first)));
        }
        m_setCount += elements.size();
    }

    /**
    @brief  Returns the number of elements in the disjoint set forest.

    @return As described
    */
    int element_count() const
    {
        return static_cast<int>(m_elements.size());
    }

    /**
    @brief  Finds the index of the root element of the tree containing x in the disjoint set forest.

    @param[in]  x   The element whose set to determine
    @pre
        -   x must be an element in the disjoint set forest
    @throw Exception
        -   If the precondition is violated
    @return As described
    */
    int find_set(int x) const
    {
        Element& element = get_element(x);
        int& parent = element.m_parent;
        if(parent != x)
        {
            parent = find_set(parent);
        }
        return parent;
    }

    /**
    @brief  Returns the current number of disjoint sets in the forest (i.e. the current number of trees).

    @return As described
    */
    int set_count() const
    {
        return m_setCount;
    }

    /**
    @brief  Merges the disjoint sets containing elements x and y.

    If both elements are already in the same disjoint set, this is a no-op.

    @param[in]  x   The first element
    @param[in]  y   The second element
    @pre
        -   Both x and y must be elements in the disjoint set forest
    @throw Exception
        -   If the precondition is violated
    */
    void union_sets(int x, int y)
    {
        int setX = find_set(x);
        int setY = find_set(y);
        if(setX != setY) link(setX, setY);
    }

    /**
    @brief  Returns the value associated with element x.

    @param[in]  x   The element whose value to return
    @pre
        -   x must be an element in the disjoint set forest
    @throw Exception
        -   If the precondition is violated
    @return As described
    */
    T& value_of(int x)
    {
        return get_element(x).m_value;
    }

    /**
    @brief  Returns the value associated with element x.

    @param[in]  x   The element whose value to return
    @pre
        -   x must be an element in the disjoint set forest
    @throw Exception
        -   If the precondition is violated
    @return As described
    */
    const T& value_of(int x) const
    {
        return get_element(x).m_value;
    }

    //#################### PRIVATE METHODS ####################
private:
    Element& get_element(int x) const
    {
        typename std::map<int,Element>::iterator it = m_elements.find(x);
        if(it != m_elements.end()) return it->second;
        else throw Exception(OSSWrapper() << "No such element: " << x);
    }

    void link(int x, int y)
    {
        Element& elementX = get_element(x);
        Element& elementY = get_element(y);
        int& rankX = elementX.m_rank;
        int& rankY = elementY.m_rank;
        if(rankX > rankY)
        {
            elementY.m_parent = x;
        }
        else
        {
            elementX.m_parent = y;
            if(rankX == rankY) ++rankY;
        }
        --m_setCount;
    }
};

}

#endif

答案 1 :(得分:3)

答案 2 :(得分:3)

你的实现看起来不错(除了在函数合并中你应该声明返回void或者返回那里,我更喜欢返回void)。 问题是你需要跟踪Nodes*。您可以通过在vector<DisjointSet::Node*>课程上设置DisjointSet或在其他地方设置vector并将DisjointSet的方法声明为static来实现此目的。

以下是run的示例(请注意,我更改了merge以返回void并且未将DisjointSet的方法更改为static

int main()
{
    vector<DisjointSet::Node*> v(10);
    DisjointSet ds;

    for (int i = 0; i < 10; ++i) {
        v[i] = new DisjointSet::Node();
        v[i]->data = i;
    }

    int x, y;

    while (cin >> x >> y) {
        ds.merge(v[x], v[y]);
    }

    for (int i = 0; i < 10; ++i) {
        cout << v[i]->data << ' ' << v[i]->parent->data << endl;
    }

    return 0;
}

使用此输入:

3 1
1 2
2 4
0 7
8 9

打印预期的:

0 7
1 1
2 1
3 1
4 1
5 5
6 6
7 7
8 9
9 9

你的森林是树木的组成:

   7    1       5    6   9
 /    / | \              |
0    2  3  4             8

所以你的算法很好,就我所知道而言具有Union-find的最佳复杂性,你跟踪Node上的vector。所以你可以简单地解除分配:

for (int i = 0; i < int(v.size()); ++i) {
    delete v[i];
}

答案 3 :(得分:2)

我不能谈论算法,但对于内存管理,通常你会使用一种称为智能指针的东西,它会释放它所指向的内容。您可以获得共享所有权和单一所有权智能指针,也可以获得非所有权。正确使用这些将保证没有内存问题。

答案 4 :(得分:0)

您的实施很好。您现在需要做的就是保留一组不相交的节点,以便您可以在它们上调用union / find方法。

对于Kruskal算法,您需要一个数组,每个图顶点包含一个不相交的节点。然后,当您查找要添加到子图的下一条边时,您将使用find方法检查这些节点是否都已存在于子图中。如果是,那么你可以继续前进到下一个边缘。否则,是时候将该边添加到子图中,并在由该边连接的两个顶点之间执行并集操作。

答案 5 :(得分:0)

这篇博客文章展示了使用路径压缩的C ++实现: http://toughprogramming.blogspot.com/2013/04/implementing-disjoint-sets-in-c.html

答案 6 :(得分:0)

看一下这段代码

class Node {
    int id,rank,data;
    Node *parent;

public :

    Node(int id,int data) {
        this->id = id;
        this->data = data;
        this->rank =0;
        this->parent = this;
    }

    friend class DisjointSet;
};

class DisjointSet {
    unordered_map<int,Node*> forest;

    Node *find_set_helper(Node *aNode) {
        if( aNode->parent != aNode)
            aNode->parent = find_set_helper(aNode->parent);

        return aNode->parent;
    }

    void link(Node *xNode,Node *yNode) {
        if( xNode->rank > yNode->rank)
            yNode->parent = xNode;
        else if(xNode-> rank < yNode->rank)
            xNode->parent = yNode;
        else {
            xNode->parent = yNode;
            yNode->rank++;
        }
    }

public:
    DisjointSet() {

    }

    void make_set(int id,int data) {
        Node *aNode = new Node(id,data);
        this->forest.insert(make_pair(id,aNode));
    }

    void Union(int xId, int yId) {
        Node *xNode = find_set(xId);
        Node *yNode = find_set(yId);

        if(xNode && yNode)
            link(xNode,yNode);
    }

    Node* find_set(int id) {
        unordered_map<int,Node*> :: iterator itr = this->forest.find(id);
        if(itr == this->forest.end())
            return NULL;

        return this->find_set_helper(itr->second);
    }

    void print() {
        unordered_map<int,Node*>::iterator itr;
        for(itr = forest.begin(); itr != forest.end(); itr++) {
            cout<<"\nid : "<<itr->second->id<<"  parent :"<<itr->second->parent->id;
        }
    }

    ~DisjointSet(){
        unordered_map<int,Node*>::iterator itr;
        for(itr = forest.begin(); itr != forest.end(); itr++) {
            delete (itr->second);
        }
    }

};

答案 7 :(得分:0)

为了从头开始实现Disjoint Sets,我强烈建议您阅读 Mark A. Weiss Data Structures & Algorithm Analysis in C++ 一书。

在第8章中,它从基本的find / union开始,然后逐渐按高度/深度/等级添加联合,并找到压缩。最后,它提供了Big-O分析。

相信我,它拥有你想要了解的关于Disjoint Sets及其C ++实现的所有内容。

答案 8 :(得分:0)

以下代码似乎很容易理解为通过路径压缩实现union-find disjoints集

$scope.myData

答案 9 :(得分:0)

如果你试图询问哪种风格更适合于不正确的不相交集(矢量或地图(rb树)),那么我可能需要添加一些内容

  1. make_set (int key , node info ):这通常是一个成员函数,用于(1)添加节点和(2)使节点指向自身(parent = key),这最终会创建一个不相交的集合。向量O(n)的操作时间复杂度,对于映射O(n * logn)。
  2. find_set( int key ):这通常有两个功能,(1)通过给定的键(2)路径压缩找到节点。我无法真正计算路径压缩,但是为了简单地搜索节点,(1)向量O(1)和(2)映射O(log(n))的时间复杂度。
  3. 最后,我想说,虽然在这里看,矢量实现看起来更好,两者的时间复杂度是O(M *α(n))≈O(M * 5)左右我读过。

    PS。尽管我确信它是正确的,但要验证我写的是什么。