Question

给定一个加权的，连通的，简单的无向图G，每条边上的权重仅为1和2

我想以这种方式实施Prim's algorithm：

权重是1或2，所以我可以简单地将边缘存储在2个单独的列表中，一个用于权重为1的边缘，第二个用于权重为2的边缘。

要找到权重最低的边，我只需从第一个列表中取一个，除非它是空的，在这种情况下，我从第二个列表中取一个边。

访问和删除列表中的元素是O（1），因此Prim的算法将在O（V + E）中运行。

package il.ac.oranim.alg2016;

import edu.princeton.cs.algs4.*; 

public class MST12 {    
    private int weight; // weight of the tree
    private Edge[] mstEdges; // use this to store the edges of your Minimum Spanning Tree

    public MST12(EdgeWeightedGraph G, int s)  throws IndexOutOfBoundsException, DisconnectedGraphException, WrongWeightException {
        // check that the starting vertex is in the range 0,1,...,G.V()
        if (s < 0 || s >= G.V()) {
            throw new IndexOutOfBoundsException();
        }
        // check that the input graph is connected otherwise there is no (minimum) spanning tree
        if (isConnected(G) == false) {
            throw new DisconnectedGraphException();
        }
        // check that all the weights are 1 or 2
        for (Edge e : G.edges()) {
            if (e.weight() != 1 && e.weight() != 2) {
                throw new WrongWeightException();
            }
        }

        this.weight = 0; // make sure you update this value

        // replace -->
        // your code goes here
        // <-- replace
    }

    // returns the weight of the tree
    public int weight() {
        return this.weight;
    }

    // checks whether a graph is connected
    private static boolean isConnected(EdgeWeightedGraph G) {
        // create a graph of class Graph with the same edges (weights)
        Graph g = new Graph(G.V());
        for (Edge e : G.edges()) {
            int v = e.either();
            g.addEdge(v, e.other(v));
        }
        // compute the connected components of the graph
        CC cc = new CC(g);

        // return true iff there is only one connected component
        return cc.count() == 1;
    }

    /**
     * Returns the edges in a minimum spanning tree as
     *    an iterable of edges
     */
    public Iterable<Edge> edges() {
        Queue<Edge> edges = new Queue<Edge>();
        for (int i = 0; i < this.mstEdges.length; i++) {
            Edge e = this.mstEdges[i];
            int v = e.either();
            edges.enqueue(new Edge(v, e.other(v), e.weight()));
        }
        return edges;
    }

    /**
     * test the computing of an MST of a graph with weights 1 and 2 only
     * the first argument is the name of the file that contains the graph (graph1.txt, graph2.txt, or graph3.txt)
     * you can define this argument in Run.. --> (x)=Arguments
     */
    public static void main(String[] args) {
        In in = new In(args[0]);
        EdgeWeightedGraph G = new EdgeWeightedGraph(in);

        PrimMST primMST = new PrimMST(G);       
        MST12 mst12 = null;
        try {
            mst12 = new MST12(G,0);
        }
        catch (DisconnectedGraphException e) {
            System.err.println("the input graph is not connected and hence has no (minimum) spanning tree");
        }
        catch (WrongWeightException e) {
            System.err.println("not all weights in the input graph are 1 or 2");            
        }

        System.out.println("Prim's MST weight = " + primMST.weight());
        System.out.println("My MST's weight = " + mst12.weight());
    }
}

我被困在//replace-->//your code goes here//replace<--
的部分

需要的两个类：

package il.ac.oranim.alg2016;

public class DisconnectedGraphException extends Exception {
    public DisconnectedGraphException() {}
}

和

  package il.ac.oranim.alg2016;

  public class WrongWeightException extends Exception {
      public WrongWeightException() {}
  }

我也允许使用所有http://algs4.cs.princeton.edu/code/

有人可以通过这部分//replace-->//your code goes here//replace<--
来帮助我

我尝试将This代码复制到//<--relpace,//replace-->部分，然后将其复制到其中，将其从使用堆更改为两个列表。

Pseudocode of Prim's algorithm

换句话说，我需要代码：

Answer 1

首先，使用在O（| E | log（| V |））中运行的优先级队列实现普通的Prim算法。并自己动手而不是复制书的代码。如果您不能自己实现Prim算法，那么您无法理解如何扩展到算法。

然后，作为D.W.在https://cs.stackexchange.com/questions/66498/prims-algorithm-on-graph-with-weights-of-only-1-and-2-on-each-edge建议您可以将ExtractMin，Remove和Insert函数更改为O（1）。

这个想法是你可以保留权重1和2的边缘列表。如果边权重1的列表不为空，那么你可以通过弹出O中的列表来获得下一个最佳边缘（1 ）时间。如果边权重1的列表为空，则可以通过在O（1）时间内弹出权重列表2来获得下一个最佳边。

普通Prim算法的唯一变化是你需要的是这样的数据结构：

private class SpecialPQ {
    ArrayList<NodeWeightPair> _queueWeight1 = new ArrayList<NodeWeightPair>();
    ArrayList<NodeWeightPair> _queueWeight2 = new ArrayList<NodeWeightPair>();

    public void insert(NodeWeightPair e) {
        if (e._weight == 1) {
            _queueWeight1.add(e);
        }
        else {
            _queueWeight2.add(e);
        }
    }

    public void remove() {
        if (_queueWeight1.size() == 0) {
            _queueWeight2.remove(_queueWeight2.size()-1);
        }
        else {
            _queueWeight1.remove(_queueWeight1.size()-1);
        }
    }

    public NodeWeightPair extractMin() {
        if (_queueWeight1.size() > 0) {
            return _queueWeight1.get(_queueWeight1.size()-1);
        }
        else {
            return _queueWeight2.get(_queueWeight2.size()-1);
        }
    }

    public boolean empty() {
        return _queueWeight1.size() == 0 && _queueWeight2.size() == 0;
    }
};

Normal Prim的算法使用二进制堆优先级队列来获得O（| E | log（| V |））。您只需用更快的SpecialPQ替换二进制堆优先级队列。

所以这本书的代码有这一行：

private IndexMinPQ<Double> pq;

您只需将其更改为

即可

private SpecialPQ pq;

并获取其余代码进行编译。不要真正复制和粘贴SpecialPQ的代码。这需要你花很长时间才能让它与书中的代码兼容。相反，我认为您应该编写自己的SpecialPQ，它将与您自己的Prim算法实现一起使用。

我在本地有一个工作示例 - 我自己的实现，因此它与本书的代码不兼容。如果您发布尝试执行此操作，我会与我分享。

编辑：

NodeWeightPair

private class NodeWeightPair {

    private int _parent;
    private int _node;
    private int _weight;
    public NodeWeightPair(int parent, int node, int weight) {
        _node = node;
        _weight = weight;
        _parent = parent;
    }
}

Answer 2

  package il.ac.oranim.alg2016;

 import edu.princeton.cs.algs4.*; 

 public class MST_12
{   
private int weight; // weight of the tree
private Edge[] mstEdges; // MST edges

private boolean[] marked;// MST vertices
private Queue<Edge> queueWeight1; 
private Queue<Edge> queueWeight2; 

   public MST_12(EdgeWeightedGraph G, int s)  throws      IndexOutOfBoundsException, DisconnectedGraphException, WrongWeightException 
{
    // check that the starting vertex is in the range 0,1,...,G.V()
    if (s < 0 || s >= G.V()) {
        throw new IndexOutOfBoundsException();
    }
    // check that the input graph is connected otherwise there is no (minimum) spanning tree
    if (isConnected(G) == false) {
        throw new DisconnectedGraphException();
    }
    // check that all the weights are 1 or 2
    for (Edge e : G.edges()) {
        if (e.weight() != 1 && e.weight() != 2) {
            throw new WrongWeightException();
        }
    }

    this.weight = 0; // make sure you update this value

    // replace -->

    queueWeight1 = new Queue<Edge>();
    queueWeight2 = new Queue<Edge>();
    mstEdges=new Edge[G.V()];
    marked=new boolean[G.V()];  

     for (int v = 0; v < G.V(); v++)      // run from each vertex to find
            if (!marked[v]) KPrim(G,v);// minimum spanning forest
}

  private void KPrim ( EdgeWeightedGraph G, int s)
    {
         visit(G,s);
         while (!queueWeight1.isEmpty()||!queueWeight2.isEmpty()){   
              Edge e=null;
              if (!queueWeight1.isEmpty())
                 { e=queueWeight1.dequeue();}
              else if (!queueWeight2.isEmpty()){e=queueWeight2.dequeue();}
              int v=e.either(), w=e.other(v);
              assert marked [v]||marked [w];
              if(marked[v]&&marked[w]) continue;
              mstEdges[s]=e;
              weight+=e.weight();
              if(!marked[v]) visit(G,v);// v becomes part of tree
              if(!marked[w]) visit(G,w);// w becomes part of a tree
         }          
    }

//add all edges e incident to v onto queue if the other endpoint has not yet been scanned
    private void visit (EdgeWeightedGraph G, int v)
    {
    marked[v]=true;// add v to T
    for (Edge e : G.adj(v))// for each edge e=v-w, add to queueWeight if w not already in T  
    { 
       if(!marked[e.other(v)])  {       
            if (e.weight()==1.0) {queueWeight1.enqueue(e);mstEdges[v]=e;}//add the smallest edge weight to the mst weight
            else {queueWeight2.enqueue(e);mstEdges[v]=e;}}}

    }

    // <-- replace


// returns the weight of the tree
public int weight() {
    return this.weight;
}

// checks whether a graph is connected
private static boolean isConnected(EdgeWeightedGraph G) {
    // create a graph of class Graph with the same edges (weights)
    Graph g = new Graph(G.V());
    for (Edge e : G.edges()) {
        int v = e.either();
        g.addEdge(v, e.other(v));
    }
    // compute the connected components of the graph
    CC cc = new CC(g);

    // return true iff there is only one connected component
    return cc.count() == 1;
}

/**
 * Returns the edges in a minimum spanning tree as
 *    an iterable of edges
 */
public Iterable<Edge> edges() {
    Queue<Edge> edges = new Queue<Edge>();
    for (int i = 0; i < this.mstEdges.length; i++) {
        Edge e = this.mstEdges[i];
        int v = e.either();
        edges.enqueue(new Edge(v, e.other(v), e.weight()));
    }
    return edges;
}

/**
 * test the computing of an MST of a graph with weights 1 and 2 only
 * the first argument is the name of the file that contains the graph (graph1.txt, graph2.txt, or graph3.txt)
 * you can define this argument in Run.. --> (x)=Arguments
 */
public static void main(String[] args) {
    In in = new In(args[0]);
    EdgeWeightedGraph G = new EdgeWeightedGraph(in);

    PrimMST primMST = new PrimMST(G);       
    MST_12 mst12 = null;
    try {
        mst12 = new MST_12(G,0);
    }
    catch (DisconnectedGraphException e) {
        System.err.println("the input graph is not connected and hence has no (minimum) spanning tree");
    }
    catch (WrongWeightException e) {
        System.err.println("not all weights in the input graph are 1 or 2");            
    }

    System.out.println("Prim's MST weight = " + primMST.weight());
    System.out.println("My MST's weight = " + mst12.weight());
}
}

使用两个列表在每个边缘上权重仅为1和2的图上的Prim算法

2 个答案: