Dijkstra 算法是否会跟踪访问过的节点以找到最短路径？

Question

我发现此代码用于Dijkstra algorithm在加权图中找到两个节点之间的最短路径。我看到的是代码没有跟踪访问过的节点。但是它适用于我尝试过的所有输入。我添加了一行代码来跟踪访问过的节点。它仍然可以正常工作。我已经在这段代码中注释掉了。那么是否需要访问节点？有没有影响O

import java.util.PriorityQueue;
import java.util.List;
import java.util.ArrayList;
import java.util.Collections;

class Vertex implements Comparable<Vertex>
{
    public final String name;
    public Edge[] adjacencies;
    public double minDistance = Double.POSITIVE_INFINITY;
    public Vertex previous;
    public Vertex(String argName) { name = argName; }
    public String toString() { return name; }
    public int compareTo(Vertex other)
    {
        return Double.compare(minDistance, other.minDistance);
    }
}

class Edge
{
    public final Vertex target;
    public final double weight;
    public Edge(Vertex argTarget, double argWeight)
    { target = argTarget; weight = argWeight; }
}

public class Dijkstra
{
    public static void computePaths(Vertex source)
    {
        source.minDistance = 0.;
        PriorityQueue<Vertex> vertexQueue = new PriorityQueue<Vertex>();
        //Set<Vertex> visited = new HashSet<Vertex>();
        vertexQueue.add(source);

    while (!vertexQueue.isEmpty()) {
        Vertex u = vertexQueue.poll();

            // Visit each edge exiting u
            for (Edge e : u.adjacencies)
            {
                Vertex v = e.target;
                double weight = e.weight;
                double distanceThroughU = u.minDistance + weight;
           //if (!visited.contains(u)){
        if (distanceThroughU < v.minDistance) {
            vertexQueue.remove(v);
            v.minDistance = distanceThroughU ;
            v.previous = u;
            vertexQueue.add(v);
                visited.add(u)
        //}
            }
        }
    }
}

    public static List<Vertex> getShortestPathTo(Vertex target)
    {
        List<Vertex> path = new ArrayList<Vertex>();
        for (Vertex vertex = target; vertex != null; vertex = vertex.previous)
            path.add(vertex);
        Collections.reverse(path);
        return path;
    }

    public static void main(String[] args)
    {
        Vertex v0 = new Vertex("Redvile");
    Vertex v1 = new Vertex("Blueville");
    Vertex v2 = new Vertex("Greenville");
    Vertex v3 = new Vertex("Orangeville");
    Vertex v4 = new Vertex("Purpleville");

    v0.adjacencies = new Edge[]{ new Edge(v1, 5),
                                 new Edge(v2, 10),
                               new Edge(v3, 8) };
    v1.adjacencies = new Edge[]{ new Edge(v0, 5),
                                 new Edge(v2, 3),
                                 new Edge(v4, 7) };
    v2.adjacencies = new Edge[]{ new Edge(v0, 10),
                               new Edge(v1, 3) };
    v3.adjacencies = new Edge[]{ new Edge(v0, 8),
                                 new Edge(v4, 2) };
    v4.adjacencies = new Edge[]{ new Edge(v1, 7),
                               new Edge(v3, 2) };
    Vertex[] vertices = { v0, v1, v2, v3, v4 };
        computePaths(v0);
        for (Vertex v : vertices)
    {
        System.out.println("Distance to " + v + ": " + v.minDistance);
        List<Vertex> path = getShortestPathTo(v);
        System.out.println("Path: " + path);
    }
    }
}

Answer 1

代码本来可以更简单，但无论如何，Djikstra 是贪婪的，所以在每个节点上，我们尝试找到路径最短的节点。除非有负边，否则已经访问过的节点已经用最短路径填充，所以很自然地，条件 if (distanceThroughU < v.minDistance) 对于访问过的节点永远不会成立。

关于运行时复杂性，您的两个实现之间不会有太大区别。