Printing Paths in Dijkstra’s Shortest Path Algorithm

• Difficulty Level : Medium
• Last Updated : 14 Sep, 2021

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 the complete implementation

C++

 // C program for Dijkstra's single // source shortest path algorithm.// The program is for adjacency matrix// representation of the graph.#include #include    // 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 treeint 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 arrayvoid 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// arrayint 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);    }}   // Function that implements Dijkstra's// single source shortest path// algorithm for a graph represented// using adjacency matrix representationvoid dijkstra(int graph[V][V], int src){           // The output array. dist[i]    // will hold the shortest    // distance from src to i    int dist[V];        // 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 = -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 Codeint 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 graphclass 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):                 #Base Case : If j is source        if parent[j] == -1 :            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)         # 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 solutiong.dijkstra(graph,0)  # This code is contributed by Neelam Yadav

Javascript



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