Dijkstra algoritme er en af de fremtrædende algoritmer til at finde den korteste vej fra kildenoden til en destinationsknuden. Den bruger den grådige tilgang til at finde den korteste vej. Konceptet med Dijkstra-algoritmen er at finde den korteste afstand (sti) fra kildepunktet og at ignorere de længere afstande, mens du laver en opdatering.
I dette afsnit vil vi implementere Dijkstra algoritme i Java-program . Vi vil også diskutere dets brug og begrænsninger.
Dijkstra Algoritme trin
Trin 1: Alle noder skal markeres som ubesøgte.
Trin 2: Alle noder skal initialiseres med den 'uendelige' (et stort tal) afstand. Startknudepunktet skal initialiseres med nul.
Trin 3: Markér startknudepunktet som den aktuelle knude.
Trin 4: Fra den aktuelle knude skal du analysere alle dens naboer, der endnu ikke er besøgt, og beregne deres afstande ved at tilføje vægten af kanten, som etablerer forbindelsen mellem den aktuelle knude og naboknude til den aktuelle afstand af den aktuelle knude.
Trin 5: Sammenlign nu den nyligt beregnede afstand med den afstand, der er tildelt den tilstødende knude, og behandl den som den aktuelle afstand til den tilstødende knude,
Trin 6: Herefter tages der hensyn til den nuværende knudes omkringliggende naboer, som ikke er besøgt, og de nuværende knudepunkter markeres som besøgt.
Trin 7: Når afslutningsknuden er markeret som besøgt, så har algoritmen gjort sit arbejde; Ellers,
Trin 8: Vælg den ubesøgte knude, som har fået tildelt den mindste afstand, og behandl den som den nye aktuelle knude. Start derefter igen fra trin 4.
Dijkstra Algorithm Pseudo Code
Method Dijkstra(G, s): // G is graph, s is source distance[s] -> 0 // Distance from the source to source is always 0 for every vertex vx in the Graph G: // doing the initialization work { if vx ? s { // Unknown distance function from source to each node set to infinity distance[vx] -> infinity } add vx to Queue Q // Initially, all the nodes are in Q } // The while loop Untill the Q is not empty: { // During the first run, this vertex is the source or starting node vx = vertex in Q with the minimum distance[vx] delete vx from Q } // where the neighbor ux has not been deleted yet from Q. for each neighbor ux of vx: alt = distance[vx] + length(vx, ux) // A path with lesser weight (shorter path), to ux is found if alt <distance[ux]: distance[ux]="alt" updating the distance of ux return dist[] end method < pre> <h2>Implementation of Dijkstra Algorithm</h2> <p>The following code implements the Dijkstra Algorithm using the diagram mentioned below.</p> <img src="//techcodeview.com/img/java-tutorial/65/dijkstra-algorithm-java.webp" alt="Dijkstra Algorithm Java"> <p> <strong>FileName:</strong> DijkstraExample.java</p> <pre> // A Java program that finds the shortest path using Dijkstra's algorithm. // The program uses the adjacency matrix for the representation of a graph // import statements import java.util.*; import java.io.*; import java.lang.*; public class DijkstraExample { // A utility method to compute the vertex with the distance value, which is minimum // from the group of vertices that has not been included yet static final int totalVertex = 9; int minimumDistance(int distance[], Boolean spSet[]) { // Initialize min value int m = Integer.MAX_VALUE, m_index = -1; for (int vx = 0; vx <totalvertex; 0 1 3 4 5 6 9 vx++) { if (spset[vx]="=" false && distance[vx] <="m)" m="distance[vx];" m_index="vx;" } return m_index; a utility method to display the built distance array void printsolution(int distance[], int n) system.out.println('the shortest from source 0th node all other nodes are: '); for (int j="0;" n; j++) system.out.println('to ' + is: distance[j]); that does implementation of dijkstra's path algorithm graph is being represented using adjacency matrix representation dijkstra(int graph[][], s) distance[]="new" int[totalvertex]; output distance[i] holds s spset[j] will be true vertex included in tree or finalized boolean spset[]="new" boolean[totalvertex]; initializing distances as infinite and totalvertex; distance[j]="Integer.MAX_VALUE;" itself always distance[s]="0;" compute given vertices cnt="0;" totalvertex - 1; cnt++) choose minimum set not yet processed. ux equal first iteration. spset); choosed marked it means processed spset[ux]="true;" updating value neighboring vertex. vx="0;" update only spset, there an edge vx, total weight through lesser than current (!spset[vx] graph[ux][vx] !="-1" distance[ux] distance[vx]) graph[ux][vx]; build printsolution(distance, totalvertex); main public static main(string argvs[]) * created. arr[x][y]="-" means, no any connects x y directly grph[][]="new" int[][] -1, 3, 7, -1 }, 10, 6, 2, 8, 13, 9, 4, 1, 5, }; creating object class dijkstraexample obj="new" dijkstraexample(); obj.dijkstra(grph, 0); pre> <p> <strong>Output:</strong> </p> <pre> The shortest Distance from source 0th node to all other nodes are: To 0 the shortest distance is: 0 To 1 the shortest distance is: 3 To 2 the shortest distance is: 8 To 3 the shortest distance is: 10 To 4 the shortest distance is: 18 To 5 the shortest distance is: 10 To 6 the shortest distance is: 9 To 7 the shortest distance is: 7 To 8 the shortest distance is: 7 </pre> <p>The time complexity of the above code is O(V<sup>2</sup>), where V is the total number of vertices present in the graph. Such time complexity does not bother much when the graph is smaller but troubles a lot when the graph is of larger size. Therefore, we have to do the optimization to reduce this complexity. With the help of the priority queue, we can decrease the time complexity. Observe the following code that is written for the graph depicted above.</p> <p> <strong>FileName:</strong> DijkstraExample1.java</p> <pre> // Java Program shows the implementation Dijkstra's Algorithm // Using the Priority Queue // import statement import java.util.*; // Main class DijkstraExample1 public class DijkstraExample1 { // Member variables of the class private int distance[]; private Set settld; private PriorityQueue pQue; // Total count of the vertices private int totalNodes; List<list> adjacent; // Constructor of the class public DijkstraExample1(int totalNodes) { this.totalNodes = totalNodes; distance = new int[totalNodes]; settld = new HashSet(); pQue = new PriorityQueue(totalNodes, new Node()); } public void dijkstra(List<list> adjacent, int s) { this.adjacent = adjacent; for (int j = 0; j <totalnodes; j++) { initializing the distance of every node to infinity (a large number) distance[j]="Integer.MAX_VALUE;" } adding source pque pque.add(new node(s, 0)); is always zero distance[s]="0;" while (settld.size() !="totalNodes)" terminating condition check when priority queue contains elements, return if (pque.isempty()) return; deleting that has minimum from int ux="pQue.remove().n;" whose confirmed (settld.contains(ux)) continue; we don't have call eneighbors(ux) already present in settled set. settld.add(ux); eneighbours(ux); private void eneighbours(int ux) edgedist="-1;" newdist="-1;" all neighbors vx for (int j="0;" < adjacent.get(ux).size(); current hasn't been processed (!settld.contains(vx.n)) + edgedist; new lesser cost (newdist distance[vx.n]) distance[vx.n]="newDist;" node(vx.n, distance[vx.n])); main method public static main(string argvs[]) totalnodes="9;" s="0;" representation connected edges using adjacency list by declaration class object declaring and type list<list> adjacent = new ArrayList<list>(); // Initialize list for every node for (int i = 0; i <totalnodes; 0 1 2 3 i++) { list itm="new" arraylist(); adjacent.add(itm); } adding the edges statement adjacent.get(0).add(new node(1, 3)); means to travel from node 1, one has cover units of distance it does not mean 0, we have add adjacent.get(1).add(new node(0, note that is same i.e., in both cases. similarly, added other too. node(7, 7)); node(2, 10)); node(8, 4)); adjacent.get(2).add(new node(3, 6)); node(5, 2)); 1)); adjacent.get(3).add(new node(4, 8)); 13)); adjacent.get(4).add(new 9)); adjacent.get(5).add(new node(6, 5)); adjacent.get(6).add(new adjacent.get(7).add(new adjacent.get(8).add(new creating an object class dijkstraexample1 obj="new" dijkstraexample1(totalnodes); obj.dijkstra(adjacent, s); printing shortest path all nodes source system.out.println('the :'); for (int j="0;" < obj.distance.length; j++) system.out.println(s + ' obj.distance[j]); implementing comparator interface this represents a graph implements member variables public int n; price; constructors constructor node() node(int n, price) this.n="n;" this.price="price;" @override compare(node n1, n2) if (n1.price n2.price) return 1; 0; pre> <p> <strong>Output:</strong> </p> <pre> The shortest path from the node: 0 to 0 is 0 0 to 1 is 3 0 to 2 is 8 0 to 3 is 10 0 to 4 is 18 0 to 5 is 10 0 to 6 is 9 0 to 7 is 7 0 to 8 is 7 </pre> <p>The time complexity of the above implementation is O(V + E*log(V)), where V is the total number of vertices, and E is the number of Edges present in the graph.</p> <h2>Limitations of Dijkstra Algorithm</h2> <p>The following are some limitations of the Dijkstra Algorithm:</p> <ol class="points"> <li>The Dijkstra algorithm does not work when an edge has negative values.</li> <li>For cyclic graphs, the algorithm does not evaluate the shortest path. Hence, for the cyclic graphs, it is not recommended to use the Dijkstra Algorithm.</li> </ol> <h2>Usages of Dijkstra Algorithm</h2> <p>A few prominent usages of the Dijkstra algorithm are:</p> <ol class="points"> <li>The algorithm is used by Google maps.</li> <li>The algorithm is used to find the distance between two locations.</li> <li>In IP routing also, this algorithm is used to discover the shortest path.</li> </ol> <hr></totalnodes;></list></totalnodes;></list></list></pre></totalvertex;></pre></distance[ux]:> Tidskompleksiteten af ovenstående kode er O(V2), hvor V er det samlede antal knudepunkter i grafen. En sådan tidskompleksitet generer ikke meget, når grafen er mindre, men generer meget, når grafen er af større størrelse. Derfor er vi nødt til at foretage optimeringen for at reducere denne kompleksitet. Ved hjælp af prioritetskøen kan vi mindske tidskompleksiteten. Overhold følgende kode, der er skrevet til grafen afbildet ovenfor.
Filnavn: DijkstraExample1.java
// Java Program shows the implementation Dijkstra's Algorithm // Using the Priority Queue // import statement import java.util.*; // Main class DijkstraExample1 public class DijkstraExample1 { // Member variables of the class private int distance[]; private Set settld; private PriorityQueue pQue; // Total count of the vertices private int totalNodes; List<list> adjacent; // Constructor of the class public DijkstraExample1(int totalNodes) { this.totalNodes = totalNodes; distance = new int[totalNodes]; settld = new HashSet(); pQue = new PriorityQueue(totalNodes, new Node()); } public void dijkstra(List<list> adjacent, int s) { this.adjacent = adjacent; for (int j = 0; j <totalnodes; j++) { initializing the distance of every node to infinity (a large number) distance[j]="Integer.MAX_VALUE;" } adding source pque pque.add(new node(s, 0)); is always zero distance[s]="0;" while (settld.size() !="totalNodes)" terminating condition check when priority queue contains elements, return if (pque.isempty()) return; deleting that has minimum from int ux="pQue.remove().n;" whose confirmed (settld.contains(ux)) continue; we don\'t have call eneighbors(ux) already present in settled set. settld.add(ux); eneighbours(ux); private void eneighbours(int ux) edgedist="-1;" newdist="-1;" all neighbors vx for (int j="0;" < adjacent.get(ux).size(); current hasn\'t been processed (!settld.contains(vx.n)) + edgedist; new lesser cost (newdist distance[vx.n]) distance[vx.n]="newDist;" node(vx.n, distance[vx.n])); main method public static main(string argvs[]) totalnodes="9;" s="0;" representation connected edges using adjacency list by declaration class object declaring and type list<list> adjacent = new ArrayList<list>(); // Initialize list for every node for (int i = 0; i <totalnodes; 0 1 2 3 i++) { list itm="new" arraylist(); adjacent.add(itm); } adding the edges statement adjacent.get(0).add(new node(1, 3)); means to travel from node 1, one has cover units of distance it does not mean 0, we have add adjacent.get(1).add(new node(0, note that is same i.e., in both cases. similarly, added other too. node(7, 7)); node(2, 10)); node(8, 4)); adjacent.get(2).add(new node(3, 6)); node(5, 2)); 1)); adjacent.get(3).add(new node(4, 8)); 13)); adjacent.get(4).add(new 9)); adjacent.get(5).add(new node(6, 5)); adjacent.get(6).add(new adjacent.get(7).add(new adjacent.get(8).add(new creating an object class dijkstraexample1 obj="new" dijkstraexample1(totalnodes); obj.dijkstra(adjacent, s); printing shortest path all nodes source system.out.println(\'the :\'); for (int j="0;" < obj.distance.length; j++) system.out.println(s + \' obj.distance[j]); implementing comparator interface this represents a graph implements member variables public int n; price; constructors constructor node() node(int n, price) this.n="n;" this.price="price;" @override compare(node n1, n2) if (n1.price n2.price) return 1; 0; pre> <p> <strong>Output:</strong> </p> <pre> The shortest path from the node: 0 to 0 is 0 0 to 1 is 3 0 to 2 is 8 0 to 3 is 10 0 to 4 is 18 0 to 5 is 10 0 to 6 is 9 0 to 7 is 7 0 to 8 is 7 </pre> <p>The time complexity of the above implementation is O(V + E*log(V)), where V is the total number of vertices, and E is the number of Edges present in the graph.</p> <h2>Limitations of Dijkstra Algorithm</h2> <p>The following are some limitations of the Dijkstra Algorithm:</p> <ol class="points"> <li>The Dijkstra algorithm does not work when an edge has negative values.</li> <li>For cyclic graphs, the algorithm does not evaluate the shortest path. Hence, for the cyclic graphs, it is not recommended to use the Dijkstra Algorithm.</li> </ol> <h2>Usages of Dijkstra Algorithm</h2> <p>A few prominent usages of the Dijkstra algorithm are:</p> <ol class="points"> <li>The algorithm is used by Google maps.</li> <li>The algorithm is used to find the distance between two locations.</li> <li>In IP routing also, this algorithm is used to discover the shortest path.</li> </ol> <hr></totalnodes;></list></totalnodes;></list></list> Tidskompleksiteten af ovenstående implementering er O(V + E*log(V)), hvor V er det samlede antal hjørner, og E er antallet af kanter til stede i grafen.
Begrænsninger af Dijkstra Algorithm
Følgende er nogle begrænsninger af Dijkstra-algoritmen:
- Dijkstra-algoritmen virker ikke, når en kant har negative værdier.
- For cykliske grafer evaluerer algoritmen ikke den korteste vej. For de cykliske grafer anbefales det derfor ikke at bruge Dijkstra-algoritmen.
Anvendelser af Dijkstra Algorithm
Et par fremtrædende anvendelser af Dijkstra-algoritmen er:
- Algoritmen bruges af Google maps.
- Algoritmen bruges til at finde afstanden mellem to steder.
- I IP-routing bruges denne algoritme også til at finde den korteste vej.