다익스트라 알고리즘은 소스 노드에서 대상 노드까지의 최단 경로를 찾는 대표적인 알고리즘 중 하나입니다. 최단 경로를 찾기 위해 탐욕적 접근 방식을 사용합니다. Dijkstra 알고리즘의 개념은 소스 지점에서 시작하여 가장 짧은 거리(경로)를 찾고 업데이트하는 동안 더 긴 거리를 무시하는 것입니다.

이 섹션에서는 Java 프로그램의 Dijkstra 알고리즘 . 또한 사용법과 제한 사항에 대해서도 설명합니다.

Dijkstra 알고리즘 단계

1 단계: 모든 노드는 방문하지 않은 것으로 표시되어야 합니다.

2 단계: 모든 노드는 '무한'(큰 숫자) 거리로 초기화되어야 합니다. 시작 노드는 0으로 초기화되어야 합니다.

3단계: 시작 노드를 현재 노드로 표시합니다.

4단계: 현재 노드에서 아직 방문하지 않은 모든 이웃을 분석하고 현재 노드의 현재 거리에 현재 노드와 이웃 노드 간의 연결을 설정하는 에지의 가중치를 추가하여 거리를 계산합니다.

5단계: 이제 최근 계산된 거리와 이웃 노드에 할당된 거리를 비교하여 현재 이웃 노드의 거리로 취급하면,

6단계: 그 후, 현재 노드의 방문하지 않은 주변 이웃을 고려하여 현재 노드를 방문했다고 표시한다.

7단계: 끝 노드가 방문한 것으로 표시되면 알고리즘이 작업을 완료한 것입니다. 그렇지 않으면,

8단계: 최소 거리가 할당된 방문하지 않은 노드를 선택하여 새로운 현재 노드로 처리합니다. 그 후 4단계부터 다시 시작하세요.

Dijkstra 알고리즘 의사 코드

 Method Dijkstra(G, s): // G is graph, s is source distance[s] -&gt; 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] -&gt; 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&apos;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&apos;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]:>

위 코드의 시간 복잡도는 O(V²), 여기서 V는 그래프에 존재하는 정점의 총 개수입니다. 이러한 시간 복잡도는 그래프가 작을 때는 크게 문제가 되지 않지만, 그래프의 크기가 클 때는 문제가 많습니다. 따라서 이러한 복잡성을 줄이기 위해 최적화를 수행해야 합니다. 우선순위 큐를 사용하면 시간 복잡도를 줄일 수 있습니다. 위에 묘사된 그래프를 위해 작성된 다음 코드를 살펴보세요.

파일 이름: DijkstraExample1.java

 // Java Program shows the implementation Dijkstra&apos;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>

위 구현의 시간 복잡도는 O(V + E*log(V))입니다. 여기서 V는 정점의 총 개수이고 E는 그래프에 존재하는 가장자리의 개수입니다.

Dijkstra 알고리즘의 한계

다음은 Dijkstra 알고리즘의 몇 가지 제한 사항입니다.

1. Dijkstra 알고리즘은 모서리에 음수 값이 있는 경우 작동하지 않습니다.
2. 순환 그래프의 경우 알고리즘은 최단 경로를 평가하지 않습니다. 따라서 순환 그래프의 경우 Dijkstra 알고리즘을 사용하지 않는 것이 좋습니다.

Dijkstra 알고리즘의 사용법

Dijkstra 알고리즘의 몇 가지 주요 용도는 다음과 같습니다.

1. 이 알고리즘은 Google 지도에서 사용됩니다.
2. 알고리즘은 두 위치 사이의 거리를 찾는 데 사용됩니다.
3. IP 라우팅에서도 이 알고리즘은 최단 경로를 찾는 데 사용됩니다.