Minimum Cost Path in a directed graph via given set of intermediate nodes

Last Updated : 05 Mar, 2023

Given a weighted, directed graph G, an array V[] consisting of vertices, the task is to find the Minimum Cost Path passing through all the vertices of the set V, from a given source S to a destination D.

Examples:

Input: V = {7}, S = 0, D = 6

Output: 11
Explanation:
Minimum path 0->7->5->6.
Therefore, the cost of the path = 3 + 6 + 2 = 11

Input: V = {7, 4}, S = 0, D = 6

Output: 12
Explanation:
Minimum path 0->7->4->6.
Therefore the cost of the path = 3 + 5 + 4 = 12

Approach:
To solve the problem, the idea is to use Breadth-First-Search traversal. BFS is generally used to find the Shortest Paths in the graph and the minimum distance of all nodes from Source, intermediate nodes, and Destination can be calculated by the BFS from these nodes.

Follow the steps below to solve the problem:

Initialize minSum to INT_MAX.
Traverse the graph from the source node S using BFS.
Mark each neighbouring node of the source as the new source and perform BFS from that node.
Once the destination node D is encountered, then check if all the intermediate nodes are visited or not.
If all the intermediate nodes are visited, then update the minSum and return the minimum value.
If all the intermediate nodes are not visited, then return minSum.
Mark the source as unvisited.
Print the final value of minSum obtained.

Below is the implementation of the above approach:

C++

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Stores minimum-cost of path from source
int minSum = INT_MAX;
 
// Function to Perform BFS on graph g
// starting from vertex v
void getMinPathSum(unordered_map<int,
                                 vector<pair<int,
                                             int> > >& graph,
                   vector<bool>& visited,
                   vector<int> necessary,
                   int src, int dest, int currSum)
{
    // If destination is reached
    if (src == dest) {
        // Set flag to true
        bool flag = true;
 
        // Visit all the intermediate nodes
        for (int i : necessary) {
 
            // If any intermediate node
            // is not visited
            if (!visited[i]) {
                flag = false;
                break;
            }
        }
 
        // If all intermediate
        // nodes are visited
        if (flag)
 
            // Update the minSum
            minSum = min(minSum, currSum);
        return;
    }
    else {
 
        // Mark the current node
        // visited
        visited[src] = true;
 
        // Traverse adjacent nodes
        for (auto node : graph[src]) {
 
            if (!visited[node.first]) {
 
                // Mark the neighbour visited
                visited[node.first] = true;
 
                // Find minimum cost path
                // considering the neighbour
                // as the source
                getMinPathSum(graph, visited,
                              necessary, node.first,
                              dest, currSum + node.second);
 
                // Mark the neighbour unvisited
                visited[node.first] = false;
            }
        }
 
        // Mark the source unvisited
        visited[src] = false;
    }
}
 
// Driver Code
int main()
{
    // Stores the graph
    unordered_map<int, vector<pair<int,
                                   int> > >
        graph;
    graph[0] = { { 1, 2 }, { 2, 3 }, { 3, 2 } };
    graph[1] = { { 4, 4 }, { 0, 1 } };
    graph[2] = { { 4, 5 }, { 5, 6 } };
    graph[3] = { { 5, 7 }, { 0, 1 } };
    graph[4] = { { 6, 4 } };
    graph[5] = { { 6, 2 } };
    graph[6] = { { 7, 11 } };
 
    // Number of nodes
    int n = 7;
 
    // Source
    int source = 0;
 
    // Destination
    int dest = 6;
 
    // Keeps a check on visited
    // and unvisited nodes
    vector<bool> visited(n, false);
 
    // Stores intermediate nodes
    vector<int> necessary{ 2, 4 };
 
    getMinPathSum(graph, visited, necessary,
                  source, dest, 0);
 
    // If no path is found
    if (minSum == INT_MAX)
        cout << "-1\n";
    else
        cout << minSum << '\n';
    return 0;
}

Java

// Java program to implement 
// the above approach 
import java.util.*;
 
class GFG{
     
static class pair
{
    int first, second;
     
    pair(int f, int s)
    {
        this.first = f;
        this.second = s;
    }
}
 
// Stores minimum-cost of path from source
static int minSum = Integer.MAX_VALUE;
 
// Function to Perform BFS on graph g
// starting from vertex v
static void getMinPathSum(Map<Integer, ArrayList<pair>> graph,
                          boolean[] visited,
                          ArrayList<Integer> necessary,
                          int source, int dest, int currSum)
{
     
    // If destination is reached
    if (src == dest)
    {
         
        // Set flag to true
        boolean flag = true;
 
        // Visit all the intermediate nodes
        for(int i : necessary)
        {
             
            // If any intermediate node
            // is not visited
            if (!visited[i])
            {
                flag = false;
                break;
            }
        }
 
        // If all intermediate
        // nodes are visited
        if (flag)
         
            // Update the minSum
            minSum = Math.min(minSum, currSum);
             
        return;
    }
    else
    {
         
        // Mark the current node
        // visited
        visited[src] = true;
 
        // Traverse adjacent nodes
        for(pair node : graph.get(src))
        {
            if (!visited[node.first]) 
            {
                 
                // Mark the neighbour visited
                visited[node.first] = true;
 
                // Find minimum cost path
                // considering the neighbour
                // as the source
                getMinPathSum(graph, visited,
                              necessary, node.first,
                              dest, currSum + node.second);
 
                // Mark the neighbour unvisited
                visited[node.first] = false;
            }
        }
 
        // Mark the source unvisited
        visited[src] = false;
    }
}
 
// Driver code
public static void main(String[] args)
{
     
    // Stores the graph
    Map<Integer, ArrayList<pair>> graph = new HashMap<>();
 
    for(int i = 0; i <= 6; i++)
        graph.put(i, new ArrayList<pair>());
 
    graph.get(0).add(new pair(1, 2));
    graph.get(0).add(new pair(2, 3));
    graph.get(0).add(new pair(3, 2));
    graph.get(1).add(new pair(4, 4));
    graph.get(1).add(new pair(0, 1));
    graph.get(2).add(new pair(4, 5));
    graph.get(2).add(new pair(5, 6));
    graph.get(3).add(new pair(5, 7));
    graph.get(3).add(new pair(0, 1));
    graph.get(4).add(new pair(6, 4));
    graph.get(5).add(new pair(4, 2));
    graph.get(6).add(new pair(7, 11));
 
    // Number of nodes
    int n = 7;
 
    // Source
    int source = 0;
 
    // Destination
    int dest = 6;
 
    // Keeps a check on visited
    // and unvisited nodes
    boolean[] visited = new boolean[n];
 
    // Stores intermediate nodes
    ArrayList<Integer> necessary = new ArrayList<>(
                                   Arrays.asList(2, 4));
 
    getMinPathSum(graph, visited, necessary,
                  source, dest, 0);
 
    // If no path is found
    if (minSum == Integer.MAX_VALUE)
        System.out.println(-1);
    else
        System.out.println(minSum);
}
}
 
// This code is contributed by offbeat

Python3

# Python3 Program to implement
# the above approach
  
# Stores minimum-cost of path from source
minSum = 1000000000
  
# Function to Perform BFS on graph g
# starting from vertex v
def getMinPathSum(graph, visited, necessary, 
                  source, dest, currSum):
     
    global minSum
     
    # If destination is reached
    if (src == dest):
       
        # Set flag to true
        flag = True;
  
        # Visit all the intermediate nodes
        for i in necessary:
  
            # If any intermediate node
            # is not visited
            if (not visited[i]):
                flag = False;
                break;
     
        # If all intermediate
        # nodes are visited
        if (flag):
  
            # Update the minSum
            minSum = min(minSum, currSum);
        return;
     
    else:
  
        # Mark the current node
        # visited
        visited[src] = True;
  
        # Traverse adjacent nodes
        for node in graph[src]:
             
            if not visited[node[0]]:
             
                # Mark the neighbour visited
                visited[node[0]] = True;
 
                # Find minimum cost path
                # considering the neighbour
                # as the source
                getMinPathSum(graph, visited,
                              necessary, node[0],
                              dest, currSum + node[1]);
 
                # Mark the neighbour unvisited
                visited[node[0]] = False;
         
        # Mark the source unvisited
        visited[src] = False;
 
# Driver Code
if __name__=='__main__':
     
    # Stores the graph
    graph=dict()
         
    graph[0] = [ [ 1, 2 ], [ 2, 3 ], [ 3, 2 ] ];
    graph[1] = [ [ 4, 4 ], [ 0, 1 ] ];
    graph[2] = [ [ 4, 5 ], [ 5, 6 ] ];
    graph[3] = [ [ 5, 7 ], [ 0, 1 ] ];
    graph[4] = [ [ 6, 4 ] ];
    graph[5] = [ [ 6, 2 ] ];
    graph[6] = [ [ 7, 11 ] ];
     
    # Number of nodes
    n = 7;
  
    # Source
    source = 0;
  
    # Destination
    dest = 6;
  
    # Keeps a check on visited
    # and unvisited nodes
    visited=[ False for i in range(n + 1)]
  
    # Stores intermediate nodes
    necessary = [ 2, 4 ];
  
    getMinPathSum(graph, visited, necessary,
                  source, dest, 0);
  
    # If no path is found
    if (minSum == 1000000000):
        print(-1)
    else:
        print(minSum)
 
        # This code is contributed by pratham76

C#

// C# program to implement 
// the above approach 
using System;
using System.Collections;
using System.Collections.Generic;
  
class GFG{
      
class pair
{
    public int first, second;
      
    public pair(int f, int s)
    {
        this.first = f;
        this.second = s;
    }
}
  
// Stores minimum-cost of path from source
static int minSum = 100000000;
  
// Function to Perform BFS on graph g
// starting from vertex v
static void getMinPathSum(Dictionary<int, ArrayList> graph,
                          bool[] visited, ArrayList necessary,
                          int source, int dest, int currSum)
{
      
    // If destination is reached
    if (src == dest)
    {
          
        // Set flag to true
        bool flag = true;
  
        // Visit all the intermediate nodes
        foreach(int i in necessary)
        {
              
            // If any intermediate node
            // is not visited
            if (!visited[i])
            {
                flag = false;
                break;
            }
        }
  
        // If all intermediate
        // nodes are visited
        if (flag)
          
            // Update the minSum
            minSum = Math.Min(minSum, currSum);
              
        return;
    }
    else
    {
          
        // Mark the current node
        // visited
        visited[src] = true;
  
        // Traverse adjacent nodes
        foreach(pair node in graph)
        {
            if (!visited[node.first]) 
            {
                  
                // Mark the neighbour visited
                visited[node.first] = true;
  
                // Find minimum cost path
                // considering the neighbour
                // as the source
                getMinPathSum(graph, visited,
                              necessary, node.first,
                              dest, currSum + node.second);
  
                // Mark the neighbour unvisited
                visited[node.first] = false;
            }
        }
  
        // Mark the source unvisited
        visited[src] = false;
    }
}
  
// Driver code
public static void Main(string[] args)
{
      
    // Stores the graph
    Dictionary<int, ArrayList> graph = new Dictionary<int, ArrayList>();
  
    for(int i = 0; i <= 6; i++)
        graph[i] = new ArrayList();
  
    graph[0].Add(new pair(1, 2));
    graph[0].Add(new pair(2, 3));
    graph[0].Add(new pair(3, 2));
    graph[1].Add(new pair(4, 4));
    graph[1].Add(new pair(0, 1));
    graph[2].Add(new pair(4, 5));
    graph[2].Add(new pair(5, 6));
    graph[3].Add(new pair(5, 7));
    graph[3].Add(new pair(0, 1));
    graph[4].Add(new pair(6, 4));
    graph[5].Add(new pair(4, 2));
    graph[6].Add(new pair(7, 11));
  
    // Number of nodes
    int n = 7;
  
    // Source
    int source = 0;
  
    // Destination
    int dest = 6;
  
    // Keeps a check on visited
    // and unvisited nodes
    bool[] visited = new bool[n];
  
    // Stores intermediate nodes
    ArrayList necessary = new ArrayList();
    necessary.Add(2);
    necessary.Add(4);
  
    getMinPathSum(graph, visited, necessary, source, dest, 0);
  
    // If no path is found
    if (minSum ==  100000000)
        Console.WriteLine(-1);
    else
        Console.WriteLine(minSum);
}
}
 
// This code is contributed by rutvik_56

Javascript

<script>
 
// Javascript program to implement 
// the above approach 
      
class pair
{
    constructor(f, s)
    {
        this.first = f;
        this.second = s;
    }
}
  
// Stores minimum-cost of path from source
var minSum = 100000000;
  
// Function to Perform BFS on graph g
// starting from vertex v
function getMinPathSum(graph, visited,necessary, src, dest, currSum)
{
      
    // If destination is reached
    if (src == dest)
    {
          
        // Set flag to true
        var flag = true;
  
        // Visit all the intermediate nodes
        for(var i of necessary)
        {
              
            // If any intermediate node
            // is not visited
            if (!visited[i])
            {
                flag = false;
                break;
            }
        }
  
        // If all intermediate
        // nodes are visited
        if (flag)
          
            // Update the minSum
            minSum = Math.min(minSum, currSum);
              
        return;
    }
    else
    {
          
        // Mark the current node
        // visited
        visited[src] = true;
  
        // Traverse adjacent nodes
        for(var node of graph[src])
        {
            if (!visited[node.first]) 
            {
                  
                // Mark the neighbour visited
                visited[node.first] = true;
  
                // Find minimum cost path
                // considering the neighbour
                // as the source
                getMinPathSum(graph, visited,
                              necessary, node.first,
                              dest, currSum + node.second);
  
                // Mark the neighbour unvisited
                visited[node.first] = false;
            }
        }
  
        // Mark the source unvisited
        visited[src] = false;
    }
}
  
// Driver code
 
// Stores the graph
var graph = Array.from(Array(7), ()=>Array());
 
graph[0].push(new pair(1, 2));
graph[0].push(new pair(2, 3));
graph[0].push(new pair(3, 2));
graph[1].push(new pair(4, 4));
graph[1].push(new pair(0, 1));
graph[2].push(new pair(4, 5));
graph[2].push(new pair(5, 6));
graph[3].push(new pair(5, 7));
graph[3].push(new pair(0, 1));
graph[4].push(new pair(6, 4));
graph[5].push(new pair(4, 2));
graph[6].push(new pair(7, 11));
 
// Number of nodes
var n = 7;
 
// Source
var source = 0;
 
// Destination
var dest = 6;
 
// Keeps a check on visited
// and unvisited nodes
var visited = Array(n).fill(false);
 
// Stores intermediate nodes
var necessary = [];
necessary.push(2);
necessary.push(4);
 
getMinPathSum(graph, visited, necessary, source, dest, 0);
 
// If no path is found
if (minSum ==  100000000)
    document.write(-1);
else
    document.write(minSum);
 
</script>

Output:

Time Complexity: O(N*M)
Auxiliary Space: O(N+M)

AnshulVerma1

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Easy problems on BFS

Intermediate problems on BFS

Hard Problems on BFS

Minimum Cost Path in a directed graph via given set of intermediate nodes

C++

Java

Python3

C#

Javascript

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