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java - Dijkstra 算法是否跟踪访问过的节点以找到最短路径?

转载 作者:行者123 更新时间:2023-12-02 17:22:55 27 4
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我找到了this code使用 Dijkstra 算法来查找加权图中两个节点之间的最短路径。我看到的是代码没有跟踪访问过的节点。但是它对于我尝试过的所有输入都适用。我添加了一行代码来跟踪访问过的节点。它仍然工作正常。我已经在这段代码中注释掉了。那么是否需要有访问过的节点呢?有什么影响O

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>();
//Set<Vertex> visited = new HashSet<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 (!visited.contains(u)){
if (distanceThroughU < v.minDistance) {
vertexQueue.remove(v);
v.minDistance = distanceThroughU ;
v.previous = u;
vertexQueue.add(v);
visited.add(u)
//}
}
}
}
}

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);
}
}
}

最佳答案

代码本来可以更简单,但不管怎样,Djikstra 是贪婪的,所以在每个节点,我们尝试找到具有最短路径的节点。除非存在负边,否则已经访问过的节点将已经填充有最短路径,因此自然地,条件 if (distanceThroughU < v.minDistance) 对于访问过的节点永远不会成立。

关于运行时复杂性,两种实现之间没有太大区别。

关于java - Dijkstra 算法是否跟踪访问过的节点以找到最短路径?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20057870/

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