Question

我正在研究Dijkstra的算法，我需要找到所有可能的最短路径。 Dijkstra的算法只返回一条短路径，如果另一条路径的成本相同，我想打印它。我没有想法，请帮助我。

谢谢。

这是我的算法：

public class Dijkstra {
  private static final Graph.Edge[] GRAPH = {
    new Graph.Edge("a", "b", 7),
    new Graph.Edge("a", "c", 9),
    new Graph.Edge("a", "f", 14),
    new Graph.Edge("b", "c", 10),
    new Graph.Edge("b", "d", 13),
    new Graph.Edge("c", "d", 11),
    new Graph.Edge("c", "f", 2),
    new Graph.Edge("d", "e", 6),
    new Graph.Edge("e", "f", 9),
  };
  private static final String START = "a";
  private static final String END = "e";

  public static void main(String[] args) {
    Graph g = new Graph(GRAPH);
    g.dijkstra(START);
    g.printPath(END);
    //g.printAllPaths();
  }
}

import java.io.*;
import java.util.*;

class Graph {
  private final Map<String, Vertex>
      graph; // mapping of vertex names to Vertex objects, built from a set of Edges

  /** One edge of the graph (only used by Graph constructor) */
  public static class Edge {
    public final String v1, v2;
    public final int dist;

    public Edge(String v1, String v2, int dist) {
      this.v1 = v1;
      this.v2 = v2;
      this.dist = dist;
    }
  }

  /** One vertex of the graph, complete with mappings to neighbouring vertices */
  public static class Vertex implements Comparable<Vertex> {
    public final String name;
    public int dist = Integer.MAX_VALUE; // MAX_VALUE assumed to be infinity
    public Vertex previous = null;
    public final Map<Vertex, Integer> neighbours = new HashMap<>();

    public Vertex(String name) {
      this.name = name;
    }

    private void printPath() {
      if (this == this.previous) {
        System.out.printf("%s", this.name);
      } else if (this.previous == null) {
        System.out.printf("%s(unreached)", this.name);
      } else {
        this.previous.printPath();
        System.out.printf(" -> %s(%d)", this.name, this.dist);
      }
    }

    public int compareTo(Vertex other) {
      return Integer.compare(dist, other.dist);
    }
  }

  /** Builds a graph from a set of edges */
  public Graph(Edge[] edges) {
    graph = new HashMap<>(edges.length);

    //one pass to find all vertices
    for (Edge e : edges) {
      if (!graph.containsKey(e.v1)) graph.put(e.v1, new Vertex(e.v1));
      if (!graph.containsKey(e.v2)) graph.put(e.v2, new Vertex(e.v2));
    }

    //another pass to set neighbouring vertices
    for (Edge e : edges) {
      graph.get(e.v1).neighbours.put(graph.get(e.v2), e.dist);
      //graph.get(e.v2).neighbours.put(graph.get(e.v1), e.dist); // also do this for an undirected graph
    }
  }

  /** Runs dijkstra using a specified source vertex */
  public void dijkstra(String startName) {
    if (!graph.containsKey(startName)) {
      System.err.printf("Graph doesn't contain start vertex \"%s\"\n", startName);
      return;
    }
    final Vertex source = graph.get(startName);
    NavigableSet<Vertex> q = new TreeSet<>();

    // set-up vertices
    for (Vertex v : graph.values()) {
      v.previous = v == source ? source : null;
      v.dist = v == source ? 0 : Integer.MAX_VALUE;
      q.add(v);
    }

    dijkstra(q);
  }

  /** Implementation of dijkstra's algorithm using a binary heap. */
  private void dijkstra(final NavigableSet<Vertex> q) {
    Vertex u, v;
    while (!q.isEmpty()) {

      u = q.pollFirst(); // vertex with shortest distance (first iteration will return source)
      if (u.dist == Integer.MAX_VALUE)
        break; // we can ignore u (and any other remaining vertices) since they are unreachable

      //look at distances to each neighbour
      for (Map.Entry<Vertex, Integer> a : u.neighbours.entrySet()) {
        v = a.getKey(); //the neighbour in this iteration

        final int alternateDist = u.dist + a.getValue();
        if (alternateDist < v.dist) { // shorter path to neighbour found
          q.remove(v);
          v.dist = alternateDist;
          v.previous = u;
          q.add(v);
        } else if (alternateDist == v.dist) {
          // Here I Would do something
        }
      }
    }
  }

  /** Prints a path from the source to the specified vertex */
  public void printPath(String endName) {
    if (!graph.containsKey(endName)) {
      System.err.printf("Graph doesn't contain end vertex \"%s\"\n", endName);
      return;
    }

    graph.get(endName).printPath();
    System.out.println();
  }
  /** Prints the path from the source to every vertex (output order is not guaranteed) */
  public void printAllPaths() {
    for (Vertex v : graph.values()) {
      v.printPath();
      System.out.println();
    }
  }

  public void printAllPaths2() {
    graph.get("e").printPath();
    System.out.println();
  }
}

Answer 1

查看所谓的k-shortest path algorithms。这些解决了在图中枚举第一，第二，......，第k个最短路径的问题。文献中有几种算法，例如参见this paper或Yen's algorithm。

注意，大多数算法不要求您预先指定k，即。您可以使用它们按递增顺序枚举最短路径，并在长度严格增加时停止。

Dijkstra的算法找到所有可能的最短路径

1 个答案: