2

我发现此代码用于Dijkstra algorithm在加权图中查找两个节点之间的最短路径。我看到的是代码没有跟踪访问的节点。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);
    }
    }
}
4

3 回答 3

5

代码本来可以更简单，但不管怎样，Djikstra 是贪婪的，所以在每个节点上，我们尝试找到具有最短路径的节点。除非存在负边，否则已经访问过的节点将已经填充了最短路径，因此自然地，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。

于 2013-11-18T21:23:14.603 回答
4

Dijkstra 算法不需要跟踪访问过的顶点，因为它会优先考虑总路径最短的那些。

对于没有立即连接到起始节点的顶点，在算法开始时它们被认为具有无限长的路径。一旦访问了一个顶点，它的所有邻居的总距离都会更新为到当前顶点的距离加上两者之间的旅行成本。

于 2013-11-18T21:19:40.353 回答
2

没有注释行包含负责将顶点对象添加到访问集的代码。看起来像：

(!visited.contains(u))

永远是真的:)

除此之外，您无需知道访问过的节点即可使用算法。

于 2013-11-18T21:20:21.333 回答