Алгоритм Дейкстры, который ищет все возможные кратчайшие пути

Question

Я работаю над алгоритмом Дейкстры, и мне нужно найти все возможные кратчайшие пути. Алгоритм Дейкстры возвращает только один короткий путь, если другой путь имеет одинаковую стоимость, я бы хотел его распечатать. У меня нет идей, пожалуйста, помогите мне.

спасибо.

Вот мой алгоритм:

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

Посмотрите в так называемый k-shortest path algorithms. Они решают задачу перечисления первого, второго, ..., k-го кратчайшего пути в графе. В литературе имеется несколько алгоритмов, см., Например, this paper или Yen's algorithm.

Обратите внимание, что большинство алгоритмов не требуют указания k авансом, т.е. вы можете использовать их для перечисления кратчайших путей в порядке возрастания и остановки, когда длина строго возрастает.