Printing Paths in Dijkstra’s Shortest Path Algorithm

Given a graph and a source vertex in graph, find shortest paths from source to all vertices in the given graph.

We have discussed Dijkstra’s Shortest Path algorithm in below posts.

The implementations discussed above only find shortest distances, but do not print paths. In this post printing of paths is discussed.

For example, consider below graph and source as 0,



Output should be
Vertex	       Distance         Path
0 -> 1 		 4		0 1 
0 -> 2 		 12		0 1 2 
0 -> 3 		 19		0 1 2 3 
0 -> 4 		 21		0 7 6 5 4 
0 -> 5 		 11		0 7 6 5 
0 -> 6 		 9		0 7 6 
0 -> 7 		 8		0 7 
0 -> 8 		 14		0 1 2 8

The idea is to create a separate array parent[]. Value of parent[v] for a vertex v stores parent vertex of v in shortest path tree. Parent of root (or source vertex) is -1. Whenever we find shorter path through a vertex u, we make u as parent of current vertex.

Once we have parent array constructed, we can print path using below recursive function.

void printPath(int parent[], int j)
{
    // Base Case : If j is source
    if (parent[j]==-1)
        return;

    printPath(parent, parent[j]);

    printf("%d ", j);
}

Below is complete implementation

C/C++

// A C / C++ program for Dijkstra's single source shortest
// path algorithm. The program is for adjacency matrix
// representation of the graph.
#include <stdio.h>
#include <limits.h>

// Number of vertices in the graph
#define V 9

// A utility function to find the vertex with minimum distance
// value, from the set of vertices not yet included in shortest
// path tree
int minDistance(int dist[], bool sptSet[])
{
    // Initialize min value
    int min = INT_MAX, min_index;

    for (int v = 0; v < V; v++)
        if (sptSet[v] == false && dist[v] <= min)
            min = dist[v], min_index = v;

    return min_index;
}

// Function to print shortest path from source to j
// using parent array
void printPath(int parent[], int j)
{
    // Base Case : If j is source
    if (parent[j]==-1)
        return;

    printPath(parent, parent[j]);

    printf("%d ", j);
}

// A utility function to print the constructed distance
// array
int printSolution(int dist[], int n, int parent[])
{
    int src = 0;
    printf("Vertex\t  Distance\tPath");
    for (int i = 1; i < V; i++)
    {
        printf("\n%d -> %d \t\t %d\t\t%d ", src, i, dist[i], src);
        printPath(parent, i);
    }
}

// Funtion that implements Dijkstra's single source shortest path
// algorithm for a graph represented using adjacency matrix
// representation
void dijkstra(int graph[V][V], int src)
{
    int dist[V];  // The output array. dist[i] will hold
                  // the shortest distance from src to i

    // sptSet[i] will true if vertex i is included / in shortest
    // path tree or shortest distance from src to i is finalized
    bool sptSet[V];

    // Parent array to store shortest path tree
    int parent[V];

    // Initialize all distances as INFINITE and stpSet[] as false
    for (int i = 0; i < V; i++)
    {
        parent[0] = -1;
        dist[i] = INT_MAX;
        sptSet[i] = false;
    }

    // Distance of source vertex from itself is always 0
    dist[src] = 0;

    // Find shortest path for all vertices
    for (int count = 0; count < V-1; count++)
    {
        // Pick the minimum distance vertex from the set of
        // vertices not yet processed. u is always equal to src
        // in first iteration.
        int u = minDistance(dist, sptSet);

        // Mark the picked vertex as processed
        sptSet[u] = true;

        // Update dist value of the adjacent vertices of the
        // picked vertex.
        for (int v = 0; v < V; v++)

            // Update dist[v] only if is not in sptSet, there is
            // an edge from u to v, and total weight of path from
            // src to v through u is smaller than current value of
            // dist[v]
            if (!sptSet[v] && graph[u][v] &&
                dist[u] + graph[u][v] < dist[v])
            {
                parent[v]  = u;
                dist[v] = dist[u] + graph[u][v];
            }  
    }

    // print the constructed distance array
    printSolution(dist, V, parent);
}

// driver program to test above function
int main()
{
    /* Let us create the example graph discussed above */
    int graph[V][V] = {{0, 4, 0, 0, 0, 0, 0, 8, 0},
                       {4, 0, 8, 0, 0, 0, 0, 11, 0},
                       {0, 8, 0, 7, 0, 4, 0, 0, 2},
                       {0, 0, 7, 0, 9, 14, 0, 0, 0},
                       {0, 0, 0, 9, 0, 10, 0, 0, 0},
                       {0, 0, 4, 0, 10, 0, 2, 0, 0},
                       {0, 0, 0, 14, 0, 2, 0, 1, 6},
                       {8, 11, 0, 0, 0, 0, 1, 0, 7},
                       {0, 0, 2, 0, 0, 0, 6, 7, 0}
                      };

    dijkstra(graph, 0);

    return 0;
}

Python

# Python program for Dijkstra's single source shortest
# path algorithm. The program is for adjacency matrix
# representation of the graph

from collections import defaultdict

#Class to represent a graph
class Graph:

	# A utility function to find the vertex with minimum dist value, from
	# the set of vertices still in queue
	def minDistance(self,dist,queue):
		# Initialize min value and min_index as -1
		minimum = float("Inf")
		min_index = -1
		#from the dist array,pick one which has min value and is till in queue
		for i in range(len(dist)):
			if dist[i] < minimum and i in queue:
				minimum = dist[i]
				min_index = i
		return min_index


	# Function to print shortest path from source to j
	# using parent array
	def printPath(self, parent, j):
		if parent[j] == -1 : #Base Case : If j is source
			print j,
			return
		self.printPath(parent , parent[j])
		print j,
		

	# A utility function to print the constructed distance
	# array
	def printSolution(self, dist, parent):
		src = 0 
		print("Vertex \t\tDistance from Source\tPath")
		for i in range(1, len(dist)):
			print("\n%d --> %d \t\t%d \t\t\t\t\t" % (src, i, dist[i])),
			self.printPath(parent,i)


	'''Function that implements Dijkstra's single source shortest path
	algorithm for a graph represented using adjacency matrix
	representation'''
	def dijkstra(self, graph, src):

		row = len(graph)
		col = len(graph[0])

		# The output array. dist[i] will hold the shortest distance from src to i
		# Initialize all distances as INFINITE 
		dist = [float("Inf")] * row

		#Parent array to store shortest path tree
		parent = [-1] * row

		# Distance of source vertex from itself is always 0
		dist[src] = 0 
	
		# Add all vertices in queue
		queue = []
		for i in range(row):
			queue.append(i)
			
		#Find shortest path for all vertices
		while queue:

			# Pick the minimum dist vertex from the set of vertices
        	# still in queue
			u = self.minDistance(dist,queue)	

			# remove min element 	
			queue.remove(u)

			# Update dist value and parent index of the adjacent vertices of
        	# the picked vertex. Consider only those vertices which are still in
        	# queue
			for i in range(col):
				'''Update dist[i] only if it is in queue, there is
            	an edge from u to i, and total weight of path from
            	src to i through u is smaller than current value of
            	dist[i]'''
				if graph[u][i] and i in queue:
					if dist[u] + graph[u][i] < dist[i]:
						dist[i] = dist[u] + graph[u][i]
						parent[i] = u


		# print the constructed distance array
		self.printSolution(dist,parent)

g= Graph()

graph = [[0, 4, 0, 0, 0, 0, 0, 8, 0],
		[4, 0, 8, 0, 0, 0, 0, 11, 0],
		[0, 8, 0, 7, 0, 4, 0, 0, 2],
		[0, 0, 7, 0, 9, 14, 0, 0, 0],
		[0, 0, 0, 9, 0, 10, 0, 0, 0],
		[0, 0, 4, 14, 10, 0, 2, 0, 0],
		[0, 0, 0, 0, 0, 2, 0, 1, 6],
		[8, 11, 0, 0, 0, 0, 1, 0, 7],
		[0, 0, 2, 0, 0, 0, 6, 7, 0]
		]

#Print the solution
g.dijkstra(graph,0)


#This code is contributed by Neelam Yadav

Output:

Vertex	       Distance         Path
0 -> 1 		 4		0 1 
0 -> 2 		 12		0 1 2 
0 -> 3 		 19		0 1 2 3 
0 -> 4 		 21		0 7 6 5 4 
0 -> 5 		 11		0 7 6 5 
0 -> 6 		 9		0 7 6 
0 -> 7 		 8		0 7 
0 -> 8 		 14		0 1 2 8 

This article is contributed by Aditya Goel. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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