Question

我需要帮助在Dijkstra算法上编写代码，以便使用java查找最短路径，并且只使用下面给出的这个版本：

**程序Dijkstra（G，w，r，Parent [0：n-1]，Dist）

对于v←0到n-1做

Dist [v]←∞

InTheTree [v]←。false。

ENDFOR

家长[r]←-1

Dist [r]←0

阶段←1到n-1的

选择最小化Dist [u]的顶点u，使得InTheTree [u] = .false。

InTheTree [u] = .true。 //将你添加到T

对于每个顶点v，使得uv∈E做//更新Dist [v]和

如果.not。 InTheTree [v]然后// Parent [v]数组

如果Dist [u]←w（uv）＆lt; Dist [v]然后

Dist [v] = Dist [u] + w（uv）

最近[v]←（）

家长[r]←u

ENDIF

ENDFOR

结束Dijkstra **

..................... 感谢

Answer 1

ArgoList有一个我刚刚用Google搜索的实现：

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>();
    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 (distanceThroughU < v.minDistance) {
        vertexQueue.remove(v);
        v.minDistance = distanceThroughU ;
        v.previous = u;
        vertexQueue.add(v);
    }
        }
    }
}

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);
}
}
}

用于寻找最短路径的Dijkstra算法

1 个答案: