Dijkstrin algoritam omogućuje nam pronalazak najkraćeg puta između bilo koja dva vrha grafa.
Razlikuje se od stabla minimalnog raspona jer najkraća udaljenost između dva vrha možda neće obuhvaćati sve vrhove grafa.
Kako djeluje Dijkstrin algoritam
Dijkstrin algoritam djeluje na osnovi toga da je bilo koji potput B -> Dnajkraćeg puta A -> Dizmeđu vrhova A i D ujedno i najkraći put između vrhova B i D.
Svaka je podputa najkraći put
Djikstra je ovo svojstvo koristio u suprotnom smjeru, tj. Precjenjujemo udaljenost svakog vrha od početnog vrha. Zatim posjetimo svaki čvor i njegove susjede kako bismo pronašli najkraći put do tih susjeda.
Algoritam koristi pohlepni pristup u smislu da pronalazimo sljedeće najbolje rješenje nadajući se da je krajnji rezultat najbolje rješenje za cijeli problem.
Primjer Dijkstrinog algoritma
Lakše je započeti s primjerom, a zatim razmisliti o algoritmu.
Započnite s ponderiranim grafom 
Odaberite početni vrh i dodijelite vrijednosti beskonačne staze svim ostalim uređajima 
Idite na svaki vrh i ažurirajte njegovu duljinu puta 
Ako je duljina puta susjednog vrha manja od nove duljine puta, nemojte ga ažurirati 
Izbjegavajte ažuriranje putanje duljine već posjećenih vrhova 
Nakon svake iteracije odabiremo ne posjećeni vrh s najmanje duljine puta. Dakle, biramo 5 prije 7 
Primijetimo kako se krajnjem desnom vrhu dva puta ažurira duljina puta. 
Ponavljajte dok se ne posjete svi vrhovi
Djikstrin pseudokod algoritam
Moramo održavati udaljenost puta svakog vrha. To možemo pohraniti u niz veličine v, gdje je v broj vrhova.
Također želimo biti u mogućnosti doći do najkraćeg puta, ne samo da znamo dužinu najkraćeg puta. Zbog toga svaki vrh preslikavamo na vrh koji je zadnji put ažurirao duljinu puta.
Nakon što algoritam završi, možemo se vratiti s odredišnog vrha na izvorni vrh kako bismo pronašli put.
Red za minimalni prioritet može se koristiti za učinkovito primanje vrha s najmanjom udaljenostom putanje.
function dijkstra(G, S) for each vertex V in G distance(V) <- infinite previous(V) <- NULL If V != S, add V to Priority Queue Q distance(S) <- 0 while Q IS NOT EMPTY U <- Extract MIN from Q for each unvisited neighbour V of U tempDistance <- distance(U) + edge_weight(U, V) if tempDistance < distance(V) distance(V) <- tempDistance previous(V) <- U return distance(), previous()
Šifra za Dijkstrin algoritam
Implementacija Dijkstrinog algoritma u C ++ data je u nastavku. Složenost koda može se poboljšati, ali apstrakcije su prikladne za povezivanje koda s algoritmom.
Python Java C C ++ # Dijkstra's Algorithm in Python import sys # Providing the graph vertices = ((0, 0, 1, 1, 0, 0, 0), (0, 0, 1, 0, 0, 1, 0), (1, 1, 0, 1, 1, 0, 0), (1, 0, 1, 0, 0, 0, 1), (0, 0, 1, 0, 0, 1, 0), (0, 1, 0, 0, 1, 0, 1), (0, 0, 0, 1, 0, 1, 0)) edges = ((0, 0, 1, 2, 0, 0, 0), (0, 0, 2, 0, 0, 3, 0), (1, 2, 0, 1, 3, 0, 0), (2, 0, 1, 0, 0, 0, 1), (0, 0, 3, 0, 0, 2, 0), (0, 3, 0, 0, 2, 0, 1), (0, 0, 0, 1, 0, 1, 0)) # Find which vertex is to be visited next def to_be_visited(): global visited_and_distance v = -10 for index in range(num_of_vertices): if visited_and_distance(index)(0) == 0 and (v < 0 or visited_and_distance(index)(1) <= visited_and_distance(v)(1)): v = index return v num_of_vertices = len(vertices(0)) visited_and_distance = ((0, 0)) for i in range(num_of_vertices-1): visited_and_distance.append((0, sys.maxsize)) for vertex in range(num_of_vertices): # Find next vertex to be visited to_visit = to_be_visited() for neighbor_index in range(num_of_vertices): # Updating new distances if vertices(to_visit)(neighbor_index) == 1 and visited_and_distance(neighbor_index)(0) == 0: new_distance = visited_and_distance(to_visit)(1) + edges(to_visit)(neighbor_index) if visited_and_distance(neighbor_index)(1)> new_distance: visited_and_distance(neighbor_index)(1) = new_distance visited_and_distance(to_visit)(0) = 1 i = 0 # Printing the distance for distance in visited_and_distance: print("Distance of ", chr(ord('a') + i), " from source vertex: ", distance(1)) i = i + 1
// Dijkstra's Algorithm in Java public class Dijkstra ( public static void dijkstra(int()() graph, int source) ( int count = graph.length; boolean() visitedVertex = new boolean(count); int() distance = new int(count); for (int i = 0; i < count; i++) ( visitedVertex(i) = false; distance(i) = Integer.MAX_VALUE; ) // Distance of self loop is zero distance(source) = 0; for (int i = 0; i < count; i++) ( // Update the distance between neighbouring vertex and source vertex int u = findMinDistance(distance, visitedVertex); visitedVertex(u) = true; // Update all the neighbouring vertex distances for (int v = 0; v < count; v++) ( if (!visitedVertex(v) && graph(u)(v) != 0 && (distance(u) + graph(u)(v) < distance(v))) ( distance(v) = distance(u) + graph(u)(v); ) ) ) for (int i = 0; i < distance.length; i++) ( System.out.println(String.format("Distance from %s to %s is %s", source, i, distance(i))); ) ) // Finding the minimum distance private static int findMinDistance(int() distance, boolean() visitedVertex) ( int minDistance = Integer.MAX_VALUE; int minDistanceVertex = -1; for (int i = 0; i < distance.length; i++) ( if (!visitedVertex(i) && distance(i) < minDistance) ( minDistance = distance(i); minDistanceVertex = i; ) ) return minDistanceVertex; ) public static void main(String() args) ( int graph()() = new int()() ( ( 0, 0, 1, 2, 0, 0, 0 ), ( 0, 0, 2, 0, 0, 3, 0 ), ( 1, 2, 0, 1, 3, 0, 0 ), ( 2, 0, 1, 0, 0, 0, 1 ), ( 0, 0, 3, 0, 0, 2, 0 ), ( 0, 3, 0, 0, 2, 0, 1 ), ( 0, 0, 0, 1, 0, 1, 0 ) ); Dijkstra T = new Dijkstra(); T.dijkstra(graph, 0); ) )
// Dijkstra's Algorithm in C #include #define INFINITY 9999 #define MAX 10 void Dijkstra(int Graph(MAX)(MAX), int n, int start); void Dijkstra(int Graph(MAX)(MAX), int n, int start) ( int cost(MAX)(MAX), distance(MAX), pred(MAX); int visited(MAX), count, mindistance, nextnode, i, j; // Creating cost matrix for (i = 0; i < n; i++) for (j = 0; j < n; j++) if (Graph(i)(j) == 0) cost(i)(j) = INFINITY; else cost(i)(j) = Graph(i)(j); for (i = 0; i < n; i++) ( distance(i) = cost(start)(i); pred(i) = start; visited(i) = 0; ) distance(start) = 0; visited(start) = 1; count = 1; while (count < n - 1) ( mindistance = INFINITY; for (i = 0; i < n; i++) if (distance(i) < mindistance && !visited(i)) ( mindistance = distance(i); nextnode = i; ) visited(nextnode) = 1; for (i = 0; i < n; i++) if (!visited(i)) if (mindistance + cost(nextnode)(i) < distance(i)) ( distance(i) = mindistance + cost(nextnode)(i); pred(i) = nextnode; ) count++; ) // Printing the distance for (i = 0; i < n; i++) if (i != start) ( printf("Distance from source to %d: %d", i, distance(i)); ) ) int main() ( int Graph(MAX)(MAX), i, j, n, u; n = 7; Graph(0)(0) = 0; Graph(0)(1) = 0; Graph(0)(2) = 1; Graph(0)(3) = 2; Graph(0)(4) = 0; Graph(0)(5) = 0; Graph(0)(6) = 0; Graph(1)(0) = 0; Graph(1)(1) = 0; Graph(1)(2) = 2; Graph(1)(3) = 0; Graph(1)(4) = 0; Graph(1)(5) = 3; Graph(1)(6) = 0; Graph(2)(0) = 1; Graph(2)(1) = 2; Graph(2)(2) = 0; Graph(2)(3) = 1; Graph(2)(4) = 3; Graph(2)(5) = 0; Graph(2)(6) = 0; Graph(3)(0) = 2; Graph(3)(1) = 0; Graph(3)(2) = 1; Graph(3)(3) = 0; Graph(3)(4) = 0; Graph(3)(5) = 0; Graph(3)(6) = 1; Graph(4)(0) = 0; Graph(4)(1) = 0; Graph(4)(2) = 3; Graph(4)(3) = 0; Graph(4)(4) = 0; Graph(4)(5) = 2; Graph(4)(6) = 0; Graph(5)(0) = 0; Graph(5)(1) = 3; Graph(5)(2) = 0; Graph(5)(3) = 0; Graph(5)(4) = 2; Graph(5)(5) = 0; Graph(5)(6) = 1; Graph(6)(0) = 0; Graph(6)(1) = 0; Graph(6)(2) = 0; Graph(6)(3) = 1; Graph(6)(4) = 0; Graph(6)(5) = 1; Graph(6)(6) = 0; u = 0; Dijkstra(Graph, n, u); return 0; )
// Dijkstra's Algorithm in C++ #include #include #define INT_MAX 10000000 using namespace std; void DijkstrasTest(); int main() ( DijkstrasTest(); return 0; ) class Node; class Edge; void Dijkstras(); vector* AdjacentRemainingNodes(Node* node); Node* ExtractSmallest(vector& nodes); int Distance(Node* node1, Node* node2); bool Contains(vector& nodes, Node* node); void PrintShortestRouteTo(Node* destination); vector nodes; vector edges; class Node ( public: Node(char id) : id(id), previous(NULL), distanceFromStart(INT_MAX) ( nodes.push_back(this); ) public: char id; Node* previous; int distanceFromStart; ); class Edge ( public: Edge(Node* node1, Node* node2, int distance) : node1(node1), node2(node2), distance(distance) ( edges.push_back(this); ) bool Connects(Node* node1, Node* node2) ( return ( (node1 == this->node1 && node2 == this->node2) || (node1 == this->node2 && node2 == this->node1)); ) public: Node* node1; Node* node2; int distance; ); /////////////////// void DijkstrasTest() ( Node* a = new Node('a'); Node* b = new Node('b'); Node* c = new Node('c'); Node* d = new Node('d'); Node* e = new Node('e'); Node* f = new Node('f'); Node* g = new Node('g'); Edge* e1 = new Edge(a, c, 1); Edge* e2 = new Edge(a, d, 2); Edge* e3 = new Edge(b, c, 2); Edge* e4 = new Edge(c, d, 1); Edge* e5 = new Edge(b, f, 3); Edge* e6 = new Edge(c, e, 3); Edge* e7 = new Edge(e, f, 2); Edge* e8 = new Edge(d, g, 1); Edge* e9 = new Edge(g, f, 1); a->distanceFromStart = 0; // set start node Dijkstras(); PrintShortestRouteTo(f); ) /////////////////// void Dijkstras() ( while (nodes.size()> 0) ( Node* smallest = ExtractSmallest(nodes); vector* adjacentNodes = AdjacentRemainingNodes(smallest); const int size = adjacentNodes->size(); for (int i = 0; i at(i); int distance = Distance(smallest, adjacent) + smallest->distanceFromStart; if (distance distanceFromStart) ( adjacent->distanceFromStart = distance; adjacent->previous = smallest; ) ) delete adjacentNodes; ) ) // Find the node with the smallest distance, // remove it, and return it. Node* ExtractSmallest(vector& nodes) ( int size = nodes.size(); if (size == 0) return NULL; int smallestPosition = 0; Node* smallest = nodes.at(0); for (int i = 1; i distanceFromStart distanceFromStart) ( smallest = current; smallestPosition = i; ) ) nodes.erase(nodes.begin() + smallestPosition); return smallest; ) // Return all nodes adjacent to 'node' which are still // in the 'nodes' collection. vector* AdjacentRemainingNodes(Node* node) ( vector* adjacentNodes = new vector(); const int size = edges.size(); for (int i = 0; i node1 == node) ( adjacent = edge->node2; ) else if (edge->node2 == node) ( adjacent = edge->node1; ) if (adjacent && Contains(nodes, adjacent)) ( adjacentNodes->push_back(adjacent); ) ) return adjacentNodes; ) // Return distance between two connected nodes int Distance(Node* node1, Node* node2) ( const int size = edges.size(); for (int i = 0; i Connects(node1, node2)) ( return edge->distance; ) ) return -1; // should never happen ) // Does the 'nodes' vector contain 'node' bool Contains(vector& nodes, Node* node) ( const int size = nodes.size(); for (int i = 0; i < size; ++i) ( if (node == nodes.at(i)) ( return true; ) ) return false; ) /////////////////// void PrintShortestRouteTo(Node* destination) ( Node* previous = destination; cout << "Distance from start: " id
node2 == node) ( cout << "adjacent: " id
Dijkstra's Algorithm Complexity
Time Complexity: O(E Log V)
where, E is the number of edges and V is the number of vertices.
Space Complexity: O(V)
Dijkstra's Algorithm Applications
- To find the shortest path
- In social networking applications
- In a telephone network
- To find the locations in the map
