java - 一段代码如何影响算法？

Question

抱歉，如果这是一个愚蠢或错误的问题，我在 Java 中找到了短路径算法的解决方案。这是代码:

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

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", 15),
      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();
   }
}

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);
            } 
         }
      }
   }

   /** 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();
      }
   }
}

我理解这个算法的大部分内容，但是方法 private void dijkstra(final NavigableSet<Vertex> q)以下问题让我感到困惑:

1. 由于它没有任何返回方法，其余代码如何评估它？
2. 使用 NavigableSet/TreeSet 是否比使用 PriorityQueue 更好？

而且，我还有一个关于在 Vertex 类中重写的 compareTo 方法的问题，它是如何被调用的？

谢谢

Answer 1

由于它没有任何返回方法，其余代码如何评估它？

具体做评估的部分是:

u = q.pollFirst();

影响什么的部分q.pollFirst()返回是:

if (alternateDist < v.dist) { // shorter path to neighbour found
    q.remove(v);
    v.dist = alternateDist;
    v.previous = u;
    q.add(v);
}

v节点从集合中移除，它的距离正在更新，然后它被重新添加到集合中。

被更新的距离是最重要的部分。

可能需要删除然后重新添加的节点，以便节点按新的距离值而不是旧的距离值排序。

一切的重点都是如此 q.pollFirst()返回距离最短的节点。

使用 NavigableSet/TreeSet 是否比使用 PriorityQueue 更好？

“更好”是主观的。您是在寻找速度或更好的界面，还是特定的数据结构？

据我了解，TreeSet和 PriorityQueue使用 Comparable对节点进行排序，因此从这个意义上说，它们的工作方式相似。

除此之外，TreeSet是一个集合，所以节点在集合中只能存在一次，而PriorityQueue是一个队列，可以多次插入同一个节点。

在这种情况下，一个集合似乎适用于 dijkstra 算法。

在类 Vertex 中重写的 compareTo 方法是如何被调用的？

compareTo TreeSet 内部使用函数在添加节点时对其进行排序。

将新节点与集合中已有的节点进行比较。 Vertex#compareTo提供算法来确定两个Vertex相互比较。在这种情况下，Vertex 的距离值进行比较。

这也暗示了为什么节点被删除并重新添加到 dijkstra(final NavigableSet<Vertex> q) 中的集合中功能。