Dijkstra’s shortest path algorithm using set in STL
Given a graph and a source vertex in graph, find shortest paths from source to all vertices in the given graph.
Input : Source = 0
Output :
Vertex Distance from Source
0 0
1 4
2 12
3 19
4 21
5 11
6 9
7 8
8 14
We have discussed Dijkstra’s shortest Path implementations.
The second implementation is time complexity wise better, but is really complex as we have implemented our own priority queue. STL provides priority_queue, but the provided priority queue doesn’t support decrease key and delete operations. And in Dijkstra’s algorithm, we need a priority queue and below operations on priority queue :
- ExtractMin : from all those vertices whose shortest distance is not yet found, we need to get vertex with minimum distance.
- DecreaseKey : After extracting vertex we need to update distance of its adjacent vertices, and if new distance is smaller, then update that in data structure.
Above operations can be easily implemented by set data structure of c++ STL, set keeps all its keys in sorted order so minimum distant vertex will always be at beginning, we can extract it from there, which is the ExtractMin operation and update other adjacent vertex accordingly if any vertex’s distance becomes smaller then delete its previous entry and insert new updated entry which is DecreaseKey operation.
Below is algorithm based on set data structure.
1) Initialize distances of all vertices as infinite.
2) Create an empty set. Every item of set is a pair
(weight, vertex). Weight (or distance) is used
as first item of pair as first item is by default
used to compare two pairs.
3) Insert source vertex into the set and make its
distance as 0.
4) While Set doesn't become empty, do following
a) Extract minimum distance vertex from Set.
Let the extracted vertex be u.
b) Loop through all adjacent of u and do
following for every vertex v.
// If there is a shorter path to v
// through u.
If dist[v] > dist[u] + weight(u, v)
(i) Update distance of v, i.e., do
dist[v] = dist[u] + weight(u, v)
(i) If v is in set, update its distance
in set by removing it first, then
inserting with new distance
(ii) If v is not in set, then insert
it in set with new distance
5) Print distance array dist[] to print all shortest
paths.
Below is the implementation of above idea.
C++
// Program to find Dijkstra's shortest path using STL set #include<bits/stdc++.h> using namespace std; # define INF 0x3f3f3f3f // This class represents a directed graph using // adjacency list representation class Graph { int V; // No. of vertices // In a weighted graph, we need to store vertex // and weight pair for every edge list< pair< int , int > > *adj; public : Graph( int V); // Constructor // function to add an edge to graph void addEdge( int u, int v, int w); // prints shortest path from s void shortestPath( int s); }; // Allocates memory for adjacency list Graph::Graph( int V) { this ->V = V; adj = new list< pair< int , int > >[V]; } void Graph::addEdge( int u, int v, int w) { adj[u].push_back(make_pair(v, w)); adj[v].push_back(make_pair(u, w)); } // Prints shortest paths from src to all other vertices void Graph::shortestPath( int src) { // Create a set to store vertices that are being // processed set< pair< int , int > > setds; // Create a vector for distances and initialize all // distances as infinite (INF) vector< int > dist(V, INF); // Insert source itself in Set and initialize its // distance as 0. setds.insert(make_pair(0, src)); dist[src] = 0; /* Looping till all shortest distance are finalized then setds will become empty */ while (!setds.empty()) { // The first vertex in Set is the minimum distance // vertex, extract it from set. pair< int , int > tmp = *(setds.begin()); setds.erase(setds.begin()); // vertex label is stored in second of pair (it // has to be done this way to keep the vertices // sorted distance (distance must be first item // in pair) int u = tmp.second; // 'i' is used to get all adjacent vertices of a vertex list< pair< int , int > >::iterator i; for (i = adj[u].begin(); i != adj[u].end(); ++i) { // Get vertex label and weight of current adjacent // of u. int v = (*i).first; int weight = (*i).second; // If there is shorter path to v through u. if (dist[v] > dist[u] + weight) { /* If distance of v is not INF then it must be in our set, so removing it and inserting again with updated less distance. Note : We extract only those vertices from Set for which distance is finalized. So for them, we would never reach here. */ if (dist[v] != INF) setds.erase(setds.find(make_pair(dist[v], v))); // Updating distance of v dist[v] = dist[u] + weight; setds.insert(make_pair(dist[v], v)); } } } // Print shortest distances stored in dist[] printf ( "Vertex Distance from Source\n" ); for ( int i = 0; i < V; ++i) printf ( "%d \t\t %d\n" , i, dist[i]); } // Driver program to test methods of graph class int main() { // create the graph given in above figure int V = 9; Graph g(V); // making above shown graph g.addEdge(0, 1, 4); g.addEdge(0, 7, 8); g.addEdge(1, 2, 8); g.addEdge(1, 7, 11); g.addEdge(2, 3, 7); g.addEdge(2, 8, 2); g.addEdge(2, 5, 4); g.addEdge(3, 4, 9); g.addEdge(3, 5, 14); g.addEdge(4, 5, 10); g.addEdge(5, 6, 2); g.addEdge(6, 7, 1); g.addEdge(6, 8, 6); g.addEdge(7, 8, 7); g.shortestPath(0); return 0; } |
Java
import java.io.*; import java.util.*; class Graph { private int V; private List<PriorityQueue<Node>> adj; public Graph( int V) { this .V = V; adj = new ArrayList<>(V); for ( int i = 0 ; i < V; ++i) adj.add( new PriorityQueue<>(Comparator.comparingInt(node -> node.weight))); } public void addEdge( int u, int v, int w) { adj.get(u).add( new Node(v, w)); adj.get(v).add( new Node(u, w)); } public void shortestPath( int src) { Set<Node> setds = new HashSet<>(); int [] dist = new int [V]; Arrays.fill(dist, Integer.MAX_VALUE); setds.add( new Node(src, 0 )); dist[src] = 0 ; while (!setds.isEmpty()) { Node tmp = setds.iterator().next(); setds.remove(tmp); int u = tmp.vertex; for (Node neighbor : adj.get(u)) { int v = neighbor.vertex; int weight = neighbor.weight; if (dist[v] > dist[u] + weight) { setds.remove( new Node(v, dist[v])); dist[v] = dist[u] + weight; setds.add( new Node(v, dist[v])); } } } System.out.println( "Vertex Distance from Source" ); for ( int i = 0 ; i < V; ++i) System.out.println(i + "\t\t" + dist[i]); } private static class Node { int vertex; int weight; public Node( int vertex, int weight) { this .vertex = vertex; this .weight = weight; } } } public class GFG { public static void main(String[] args) { int V = 9 ; Graph g = new Graph(V); g.addEdge( 0 , 1 , 4 ); g.addEdge( 0 , 7 , 8 ); g.addEdge( 1 , 2 , 8 ); g.addEdge( 1 , 7 , 11 ); g.addEdge( 2 , 3 , 7 ); g.addEdge( 2 , 8 , 2 ); g.addEdge( 2 , 5 , 4 ); g.addEdge( 3 , 4 , 9 ); g.addEdge( 3 , 5 , 14 ); g.addEdge( 4 , 5 , 10 ); g.addEdge( 5 , 6 , 2 ); g.addEdge( 6 , 7 , 1 ); g.addEdge( 6 , 8 , 6 ); g.addEdge( 7 , 8 , 7 ); g.shortestPath( 0 ); } } |
Python3
import heapq class Graph: def __init__( self , V): self .V = V self .adj = [[] for _ in range (V)] def addEdge( self , u, v, w): self .adj[u].append((v, w)) self .adj[v].append((u, w)) def shortestPath( self , src): # Create a priority queue to store vertices that are being processed pq = [( 0 , src)] # Create a list for distances and initialize all distances as infinite (INF) dist = [ float ( 'inf' )] * self .V dist[src] = 0 # Looping until all shortest distances are finalized while pq: # The first vertex in the priority queue is the minimum distance vertex d, u = heapq.heappop(pq) # Process the vertex if it has not been processed yet if d > dist[u]: continue # Iterate through all adjacent vertices of 'u' for v, weight in self .adj[u]: # If there is a shorter path to 'v' through 'u' if dist[v] > dist[u] + weight: # Update the distance of 'v' dist[v] = dist[u] + weight heapq.heappush(pq, (dist[v], v)) # Print shortest distances stored in dist[] print ( "Vertex Distance from Source" ) for i in range ( self .V): print (f "{i} \t\t {dist[i]}" ) # Driver program to test methods of the Graph class if __name__ = = "__main__" : # create the graph given in the above figure V = 9 g = Graph(V) # making the above shown graph g.addEdge( 0 , 1 , 4 ) g.addEdge( 0 , 7 , 8 ) g.addEdge( 1 , 2 , 8 ) g.addEdge( 1 , 7 , 11 ) g.addEdge( 2 , 3 , 7 ) g.addEdge( 2 , 8 , 2 ) g.addEdge( 2 , 5 , 4 ) g.addEdge( 3 , 4 , 9 ) g.addEdge( 3 , 5 , 14 ) g.addEdge( 4 , 5 , 10 ) g.addEdge( 5 , 6 , 2 ) g.addEdge( 6 , 7 , 1 ) g.addEdge( 6 , 8 , 6 ) g.addEdge( 7 , 8 , 7 ) g.shortestPath( 0 ) |
C#
using System; using System.Collections.Generic; using System.Linq; class Graph { private int V; // No. of vertices // In a weighted graph, we need to store vertex // and weight pair for every edge private List<Tuple< int , int > >[] adj; public Graph( int V) // Constructor { this .V = V; adj = new List<Tuple< int , int > >[ V ]; for ( int i = 0; i < V; i++) { adj[i] = new List<Tuple< int , int > >(); } } public void AddEdge( int u, int v, int w) { adj[u].Add(Tuple.Create(v, w)); adj[v].Add(Tuple.Create(u, w)); } // Prints shortest paths from src to all other vertices public void ShortestPath( int src) { // Create a set to store vertices that are being // processed SortedSet<Tuple< int , int > > setds = new SortedSet<Tuple< int , int > >(); // Create an array for distances and initialize all // distances as infinite (INF) int [] dist = Enumerable.Repeat( int .MaxValue, V).ToArray(); // Insert source itself in Set and initialize its // distance as 0. setds.Add(Tuple.Create(0, src)); dist[src] = 0; /* Looping until all shortest distances are finalized, then setds will become empty */ while (setds.Count > 0) { // The first vertex in Set is the minimum // distance vertex, extract it from set. Tuple< int , int > tmp = setds.First(); setds.Remove(tmp); // Vertex label is stored in the second item of // the tuple. int u = tmp.Item2; // Iterate through all adjacent vertices of u. foreach ( var edge in adj[u]) { int v = edge.Item1; // Get vertex label int weight = edge.Item2; // Get weight of // the current edge // If there is a shorter path to v through // u. if (dist[v] > dist[u] + weight) { /* If the distance of v is not int.MaxValue, then it must be in our set, so remove it and insert again with the updated shorter distance. Note: We extract only those vertices from the set for which the distance is finalized, so for them, we would never reach here. */ if (dist[v] != int .MaxValue) { setds.Remove( Tuple.Create(dist[v], v)); } // Update the distance of v. dist[v] = dist[u] + weight; setds.Add(Tuple.Create(dist[v], v)); } } } // Print shortest distances stored in dist[] Console.WriteLine( "Vertex Distance from Source" ); for ( int i = 0; i < V; ++i) { Console.WriteLine(i + "\t\t" + dist[i]); } } } // Driver program to test methods of the Graph class class Program { public static void Main() { // Create the graph as shown in the provided C++ // code int V = 9; Graph g = new Graph(V); g.AddEdge(0, 1, 4); g.AddEdge(0, 7, 8); g.AddEdge(1, 2, 8); g.AddEdge(1, 7, 11); g.AddEdge(2, 3, 7); g.AddEdge(2, 8, 2); g.AddEdge(2, 5, 4); g.AddEdge(3, 4, 9); g.AddEdge(3, 5, 14); g.AddEdge(4, 5, 10); g.AddEdge(5, 6, 2); g.AddEdge(6, 7, 1); g.AddEdge(6, 8, 6); g.AddEdge(7, 8, 7); g.ShortestPath(0); } } |
Javascript
// Program to find Dijkstra's shortest path using JavaScript Set const INF = 0x3f3f3f3f; // This class represents a directed graph using // adjacency list representation class Graph { constructor(V) { this .V = V; // No. of vertices // In a weighted graph, we need to store vertex // and weight pair for every edge this .adj = new Array(V); for (let i = 0; i < V; i++) { this .adj[i] = []; } } // function to add an edge to graph addEdge(u, v, w) { this .adj[u].push({ v, w }); this .adj[v].push({ u, w }); } // prints shortest path from s shortestPath(src) { // Create a set to store vertices that are being // processed const setds = new Set(); // Create an array for distances and initialize all // distances as infinite (INF) const dist = new Array( this .V).fill(INF); // Insert source itself in Set and initialize its // distance as 0. setds.add([0, src]); dist[src] = 0; /* Looping till all shortest distance are finalized then setds will become empty */ while (setds.size !== 0) { // The first vertex in Set is the minimum distance // vertex, extract it from set. const tmp = setds.values().next().value; setds. delete (tmp); // vertex label is stored in second of pair (it // has to be done this way to keep the vertices // sorted distance (distance must be first item // in pair) const u = tmp[1]; // 'i' is used to get all adjacent vertices of a vertex for (const i of this .adj[u]) { // Get vertex label and weight of current adjacent // of u. const { v, w } = i; // If there is shorter path to v through u. if (dist[v] > dist[u] + w) { /* If distance of v is not INF then it must be in our set, so removing it and inserting again with updated less distance. Note : We extract only those vertices from Set for which distance is finalized. So for them, we would never reach here. */ if (dist[v] !== INF) { setds. delete ([dist[v], v]); } // Updating distance of v dist[v] = dist[u] + w; setds.add([dist[v], v]); } } } // Print shortest distances stored in dist[] console.log( "Vertex Distance from Source" ); for (let i = 0; i < this .V; i++) { console.log(`${i} \t\t ${dist[i]}`); } } } // Driver program to test methods of graph class function main() { // create the graph given in above figure const V = 9; const g = new Graph(V); // making above shown graph g.addEdge(0, 1, 4); g.addEdge(0, 7, 8); g.addEdge(1, 2, 8); g.addEdge(1, 7, 11); g.addEdge(2, 3, 7); g.addEdge(2, 8, 2); g.addEdge(2, 5, 4); g.addEdge(3, 4, 9); g.addEdge(3, 5, 14); g.addEdge(4, 5, 10); g.addEdge(5, 6, 2); g.addEdge(6, 7, 1); g.addEdge(6, 8, 6); g.addEdge(7, 8, 7); g.shortestPath(0); } main(); |
Vertex Distance from Source 0 0 1 4 2 12 3 19 4 21 5 11 6 9 7 8 8 14
Time Complexity: Set in C++ are typically implemented using Self-balancing binary search trees. Therefore, time complexity of set operations like insert, delete is logarithmic and time complexity of above solution is O(ELogV)).
Space Complexuty:O(V2) , here V is number of Vertices.
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