Java - 在距离加权映射中查找2个点之间的最短路径

Question

我需要一种算法来查找地图中两点之间的最短路径,其中道路距离由数字表示.

给出的内容:开始城市目的地城市Z.

城市间距离列表:

A - B:10
F - K:23
R - M:8
K - O:40
Z - P:18
J - K:25
D - B:11
M - A:8
P - R:15

我想我可以使用Dijkstra的算法,但它找到了到所有目的地的最短距离.不只是一个.

任何建议表示赞赏.

Answer 1

像SplinterReality说: There's no reason not to use Dijkstra's algorithm here.

下面的代码我从这里开始修改并修改它以解决问题中的示例.

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)
    {
        // mark all the vertices 
        Vertex A = new Vertex("A");
        Vertex B = new Vertex("B");
        Vertex D = new Vertex("D");
        Vertex F = new Vertex("F");
        Vertex K = new Vertex("K");
        Vertex J = new Vertex("J");
        Vertex M = new Vertex("M");
        Vertex O = new Vertex("O");
        Vertex P = new Vertex("P");
        Vertex R = new Vertex("R");
        Vertex Z = new Vertex("Z");

        // set the edges and weight
        A.adjacencies = new Edge[]{ new Edge(M, 8) };
        B.adjacencies = new Edge[]{ new Edge(D, 11) };
        D.adjacencies = new Edge[]{ new Edge(B, 11) };
        F.adjacencies = new Edge[]{ new Edge(K, 23) };
        K.adjacencies = new Edge[]{ new Edge(O, 40) };
        J.adjacencies = new Edge[]{ new Edge(K, 25) };
        M.adjacencies = new Edge[]{ new Edge(R, 8) };
        O.adjacencies = new Edge[]{ new Edge(K, 40) };
        P.adjacencies = new Edge[]{ new Edge(Z, 18) };
        R.adjacencies = new Edge[]{ new Edge(P, 15) };
        Z.adjacencies = new Edge[]{ new Edge(P, 18) };


        computePaths(A); // run Dijkstra
        System.out.println("Distance to " + Z + ": " + Z.minDistance);
        List<Vertex> path = getShortestPathTo(Z);
        System.out.println("Path: " + path);
    }
}

上面的代码产生:

Distance to Z: 49.0
Path: [A, M, R, P, Z]

Answer 2

估计的sanjan:

Dijkstra算法背后的想法是以有序的方式探索图的所有节点.该算法存储优先级队列,其中节点根据开始时的成本排序,并且在算法的每次迭代中执行以下操作:

1. 从队列中提取从一开始就具有最低成本的节点N.
2. 获取其邻居(N')及其相关成本,即成本(N)+成本(N,N')
3. 在队列中插入邻居节点N',优先级由其成本给出

确实,该算法计算了起点(在您的情况下为A)与所有其余节点之间的路径成本,但您可以在算法到达目标时停止对该算法的探索(在您的示例中为Z).此时,您知道A和Z之间的成本以及连接它们的路径.

我建议你使用一个实现这个算法的库,而不是自己编写代码.在Java中,您可以查看一下Hipster库,它有一种非常友好的方式来生成图形并开始使用搜索算法.

这里有一个如何定义图形并开始使用Dijstra with Hipster的示例.

// Create a simple weighted directed graph with Hipster where
// vertices are Strings and edge values are just doubles
HipsterDirectedGraph<String,Double> graph = GraphBuilder.create()
  .connect("A").to("B").withEdge(4d)
  .connect("A").to("C").withEdge(2d)
  .connect("B").to("C").withEdge(5d)
  .connect("B").to("D").withEdge(10d)
  .connect("C").to("E").withEdge(3d)
  .connect("D").to("F").withEdge(11d)
  .connect("E").to("D").withEdge(4d)
  .buildDirectedGraph();

// Create the search problem. For graph problems, just use
// the GraphSearchProblem util class to generate the problem with ease.
SearchProblem p = GraphSearchProblem
  .startingFrom("A")
  .in(graph)
  .takeCostsFromEdges()
  .build();

// Search the shortest path from "A" to "F"
System.out.println(Hipster.createDijkstra(p).search("F"));

您只需要替换您自己的图形定义,然后像示例中那样实例化算法.

我希望这有帮助!