95 lines
3.4 KiB
Markdown
95 lines
3.4 KiB
Markdown
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---
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title: Dijkstra's Algorithm
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localeTitle: Dijkstra的算法
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---
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# Dijkstra的算法
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Dijkstra算法是由EW Dijkstra提出的图算法。它在具有非负边的图中找到单源最短路径。（为什么？）
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我们创建了2个数组：visit和distance，它们分别记录是否访问了顶点以及距离源顶点的最小距离。最初访问的数组被指定为false，距离指定为无限。
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我们从源顶点开始。设当前顶点为u，其相邻顶点为v。现在对于与u相邻的每个v，如果之前未访问过该距离并且距离u的距离小于其当前距离，则更新距离。然后我们选择距离最小且未访问过的下一个顶点。
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优先级队列通常用于在最短的时间内满足最后的要求。下面是使用Java中的优先级队列实现相同的想法。
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```java
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import java.util.*;
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public class Dijkstra {
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class Graph {
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LinkedList<Pair<Integer>> adj[];
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int n; // Number of vertices.
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Graph(int n) {
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this.n = n;
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adj = new LinkedList[n];
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for(int i = 0;i<n;i++) adj[i] = new LinkedList<>();
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}
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// add a directed edge between vertices a and b with cost as weight
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public void addEdgeDirected(int a, int b, int cost) {
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adj[a].add(new Pair(b, cost));
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}
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public void addEdgeUndirected(int a, int b, int cost) {
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addEdgeDirected(a, b, cost);
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addEdgeDirected(b, a, cost);
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}
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}
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class Pair<E> {
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E first;
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E second;
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Pair(E f, E s) {
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first = f;
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second = s;
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}
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}
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// Comparator to sort Pairs in Priority Queue
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class PairComparator implements Comparator<Pair<Integer>> {
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public int compare(Pair<Integer> a, Pair<Integer> b) {
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return a.second - b.second;
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}
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}
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// Calculates shortest path to each vertex from source and returns the distance
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public int[] dijkstra(Graph g, int src) {
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int distance[] = new int[gn]; // shortest distance of each vertex from src
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boolean visited[] = new boolean[gn]; // vertex is visited or not
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Arrays.fill(distance, Integer.MAX_VALUE);
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Arrays.fill(visited, false);
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PriorityQueue<Pair<Integer>> pq = new PriorityQueue<>(100, new PairComparator());
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pq.add(new Pair<Integer>(src, 0));
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distance[src] = 0;
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while(!pq.isEmpty()) {
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Pair<Integer> x = pq.remove(); // Extract vertex with shortest distance from src
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int u = x.first;
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visited[u] = true;
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Iterator<Pair<Integer>> iter = g.adj[u].listIterator();
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// Iterate over neighbours of u and update their distances
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while(iter.hasNext()) {
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Pair<Integer> y = iter.next();
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int v = y.first;
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int weight = y.second;
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// Check if vertex v is not visited
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// If new path through u offers less cost then update distance array and add to pq
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if(!visited[v] && distance[u]+weight<distance[v]) {
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distance[v] = distance[u]+weight;
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pq.add(new Pair(v, distance[v]));
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}
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}
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}
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return distance;
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}
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public static void main(String args[]) {
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Dijkstra d = new Dijkstra();
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Dijkstra.Graph g = d.new Graph(4);
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g.addEdgeUndirected(0, 1, 2);
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g.addEdgeUndirected(1, 2, 1);
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g.addEdgeUndirected(0, 3, 6);
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g.addEdgeUndirected(2, 3, 1);
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g.addEdgeUndirected(1, 3, 3);
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int dist[] = d.dijkstra(g, 0);
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System.out.println(Arrays.toString(dist));
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}
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}
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```
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