Dijkstra使用PriorityQueue的Java中最短路径算法 _JavaScript

【Dijkstra使用PriorityQueue的Java中最短路径算法】给定一个带有邻接表表示节点之间边缘的图形, 任务是实现Dijkstra的算法对于单源最短路径使用优先队列在Java中。
给定一个图和图中的一个源顶点, 找到从源到给定图中所有顶点的最短路径。

Input : Source = 0Output : VertexDistance from Source00142123194215116978814

我们已经讨论了Dijkstra最短的Path实现, 例如Dijkstra的邻接矩阵表示算法(时间复杂度为O(v2)
以下是Dijkstra算法使用优先级队列的Java实现：

//Java implementation of Dijkstra's Algorithm //using Priority Queue import java.util.*; public class DPQ { private int dist[]; private Set< Integer> settled; private PriorityQueue< Node> pq; private int V; //Number of vertices List< List< Node> > adj; public DPQ( int V) { this .V = V; dist = new int [V]; settled = new HashSet< Integer> (); pq = new PriorityQueue< Node> (V, new Node()); }//Function for Dijkstra's Algorithm public void dijkstra(List< List< Node> > adj, int src) { this .adj = adj; for ( int i = 0 ; i < V; i++) dist[i] = Integer.MAX_VALUE; //Add source node to the priority queue pq.add( new Node(src, 0 )); //Distance to the source is 0 dist[src] = 0 ; while (settled.size() != V) {//remove the minimum distance node //from the priority queue int u = pq.remove().node; //adding the node whose distance is //finalized settled.add(u); e_Neighbours(u); } }//Function to process all the neighbours //of the passed node private void e_Neighbours( int u) { int edgeDistance = - 1 ; int newDistance = - 1 ; //All the neighbors of v for ( int i = 0 ; i < adj.get(u).size(); i++) { Node v = adj.get(u).get(i); //If current node hasn't already been processed if (!settled.contains(v.node)) { edgeDistance = v.cost; newDistance = dist[u] + edgeDistance; //If new distance is cheaper in cost if (newDistance < dist[v.node]) dist[v.node] = newDistance; //Add the current node to the queue pq.add( new Node(v.node, dist[v.node])); } } }//Driver code public static void main(String arg[]) { int V = 5 ; int source = 0 ; //Adjacency list representation of the //connected edges List< List< Node> > adj = new ArrayList< List< Node> > (); //Initialize list for every node for ( int i = 0 ; i < V; i++) { List< Node> item = new ArrayList< Node> (); adj.add(item); }//Inputs for the DPQ graph adj.get( 0 ).add( new Node( 1 , 9 )); adj.get( 0 ).add( new Node( 2 , 6 )); adj.get( 0 ).add( new Node( 3 , 5 )); adj.get( 0 ).add( new Node( 4 , 3 )); adj.get( 2 ).add( new Node( 1 , 2 )); adj.get( 2 ).add( new Node( 3 , 4 )); //Calculate the single source shortest path DPQ dpq = new DPQ(V); dpq.dijkstra(adj, source); //Print the shortest path to all the nodes //from the source node System.out.println( "The shorted path from node :" ); for ( int i = 0 ; i < dpq.dist.length; i++) System.out.println(source + " to " + i + " is " + dpq.dist[i]); } }//Class to represent a node in the graph class Node implements Comparator< Node> { public int node; public int cost; public Node() { }public Node( int node, int cost) { this .node = node; this .cost = cost; }@Override public int compare(Node node1, Node node2) { if (node1.cost < node2.cost) return - 1 ; if (node1.cost> node2.cost) return 1 ; return 0 ; } }

输出如下：