Minimum Cost of Simple Path between two nodes in a Directed and Weighted Graph

Given a directed graph, which may contain cycles, where every edge has weight, the task is to find the minimum cost of any simple path from a given source vertex ‘s’ to a given destination vertex ‘t’. Simple Path is the path from one vertex to another such that no vertex is visited more than once. If there is no simple path possible then return INF(infinite).

The graph is given as adjacency matrix representation where value of graph[i][j] indicates the weight of an edge from vertex i to vertex j and a value INF(infinite) indicates no edge from i to j.

Examples:

Input : V = 5, E = 6
        s = 0, t = 2
    graph[][] =      0   1   2   3   4  
                 0  INF -1  INF  1  INF
                 1  INF INF -2  INF INF
                 2  -3  INF INF INF INF
                 3  INF INF -1  INF INF
                 4  INF INF INF  2  INF
 
Output : -3 
Explanation : 
The minimum cost simple path between 0 and 2 is given by:
0 -----> 1 ------> 2 whose cost is (-1) + (-2) = (-3). 

Input : V = 5, E = 6
        s = 0, t = 4
    graph[][] =      0   1   2   3   4  
                 0  INF -7  INF -2  INF
                 1  INF INF -11 INF INF
                 2  INF INF INF INF INF
                 3  INF INF INF  3  -4
                 4  INF INF INF INF INF
 
Output : -6
Explanation : 
The minimum cost simple path between 0 and 2 is given by:
0 -----> 3 ------> 4 whose cost is (-2) + (-4) = (-6). 

Approach :
The main idea to solve the above problem is to traverse through all simple paths from s to t using a modified version of Depth First Search and find the minimum cost path amongst them. One important observation about DFS is that it traverses one path at a time, hence we can traverse separate paths independently using DFS by marking the nodes as unvisited before leaving them.
A simple solution is to start from s, go to all adjacent vertices, and follow recursion for further adjacent vertices until we reach the destination. This algorithm will work even when negative weight cycles or self-edges are present in the graph.

Below is the implementation of the above-mentioned approach: 

C++

// C++ code for printing Minimum Cost
// Simple Path between two given nodes
// in a directed and weighted graph
#include <bits/stdc++.h>
using namespace std;

// Define number of vertices in
// the graph and infinite value
#define V 5
#define INF INT_MAX

// Function to do DFS through the nodes
int minimumCostSimplePath(int u, int destination,
                    bool visited[], int graph[][V])
{

    // check if we find the destination
    // then further cost will be 0
    if (u == destination)
        return 0;

    // marking the current node as visited
    visited[u] = 1;

    int ans = INF;

    // traverse through all
    // the adjacent nodes
    for (int i = 0; i < V; i++) {
        if (graph[u][i] != INF && !visited[i]) {

            // cost of the further path
            int curr = minimumCostSimplePath(i,
                        destination, visited, graph);

            // check if we have reached the destination
            if (curr < INF) {

                // Taking the minimum cost path
                ans = min(ans, graph[u][i] + curr);
            }
        }
    }

    // unmarking the current node
    // to make it available for other
    // simple paths
    visited[u] = 0;

    // returning the minimum cost
    return ans;
}

// driver code
int main()
{

    // initialising the graph
    int graph[V][V];
    for (int i = 0; i < V; i++) {
        for (int j = 0; j < V; j++) {
            graph[i][j] = INF;
        }
    }

    // marking all nodes as unvisited
    bool visited[V] = { 0 };

    // initialising the edges;
    graph[0][1] = -1;
    graph[0][3] = 1;
    graph[1][2] = -2;
    graph[2][0] = -3;
    graph[3][2] = -1;
    graph[4][3] = 2;

    // source and destination
    int s = 0, t = 2;

    // marking the source as visited
    visited[s] = 1;

    cout << minimumCostSimplePath(s, t,
                            visited, graph);

    return 0;
}

Java

// Java code for printing Minimum Cost
// Simple Path between two given nodes
// in a directed and weighted graph
import java.util.*;
import java.lang.*;

class GFG{

// Define number of vertices in
// the graph and infinite value
static int V = 5;
static int INF = Integer.MAX_VALUE;

// Function to do DFS through the nodes
static int minimumCostSimplePath(int u, int destination,
                                 boolean visited[],
                                 int graph[][])
{

    // Check if we find the destination
    // then further cost will be 0
    if (u == destination)
        return 0;

    // Marking the current node as visited
    visited[u] = true;

    int ans = INF;

    // Traverse through all
    // the adjacent nodes
    for(int i = 0; i < V; i++)
    {
        if (graph[u][i] != INF && !visited[i])
        {

            // Cost of the further path
            int curr = minimumCostSimplePath(i,
                        destination, visited, graph);

            // Check if we have reached the
            // destination
            if (curr < INF)
            {

                // Taking the minimum cost path
                ans = Math.min(ans, graph[u][i] + curr);
            }
        }
    }

    // Unmarking the current node
    // to make it available for other
    // simple paths
    visited[u] = false;

    // Returning the minimum cost
    return ans;
}

// Driver code
public static void main(String[] args)
{

    // Initialising the graph
    int graph[][] = new int[V][V];
    for(int i = 0; i < V; i++)
    {
        for(int j = 0; j < V; j++)
        {
            graph[i][j] = INF;
        }
    }

    // Marking all nodes as unvisited
    boolean visited[] = new boolean[V];

    // Initialising the edges;
    graph[0][1] = -1;
    graph[0][3] = 1;
    graph[1][2] = -2;
    graph[2][0] = -3;
    graph[3][2] = -1;
    graph[4][3] = 2;

    // Source and destination
    int s = 0, t = 2;

    // Marking the source as visited
    visited[s] = true;

    System.out.println(minimumCostSimplePath(s, t,
                            visited, graph));
}
}

// This code is contributed by offbeat

Python3

# Python3 code for printing Minimum Cost
# Simple Path between two given nodes
# in a directed and weighted graph
import sys

V = 5
INF = sys.maxsize

# Function to do DFS through the nodes
def minimumCostSimplePath(u, destination,
                          visited, graph):

    # Check if we find the destination
    # then further cost will be 0
    if (u == destination):
        return 0

    # Marking the current node as visited
    visited[u] = 1

    ans = INF

    # Traverse through all
    # the adjacent nodes
    for i in range(V):
        if (graph[u][i] != INF and not visited[i]):

            # Cost of the further path
            curr = minimumCostSimplePath(i, destination,
                                         visited, graph)

            # Check if we have reached the destination
            if (curr < INF):

                # Taking the minimum cost path
                ans = min(ans, graph[u][i] + curr)

    # Unmarking the current node
    # to make it available for other
    # simple paths
    visited[u] = 0

    # Returning the minimum cost
    return ans

# Driver code
if __name__=="__main__":

    # Initialising the graph
    graph = [[INF for j in range(V)]
                  for i in range(V)]

    # Marking all nodes as unvisited
    visited = [0 for i in range(V)]

    # Initialising the edges
    graph[0][1] = -1
    graph[0][3] = 1
    graph[1][2] = -2
    graph[2][0] = -3
    graph[3][2] = -1
    graph[4][3] = 2

    # Source and destination
    s = 0
    t = 2

    # Marking the source as visited
    visited[s] = 1

    print(minimumCostSimplePath(s, t, visited, graph))

# This code is contributed by rutvik_56

C#

// C# code for printing Minimum Cost
// Simple Path between two given nodes
// in a directed and weighted graph
using System;
using System.Collections;
using System.Collections.Generic;
class GFG
{

    // Define number of vertices in
    // the graph and infinite value
    static int V = 5;
    static int INF = int.MaxValue;

    // Function to do DFS through the nodes
    static int minimumCostSimplePath(int u, int destination,
                                     bool[] visited, int[, ] graph)
    {

        // Check if we find the destination
        // then further cost will be 0
        if (u == destination)
            return 0;

        // Marking the current node as visited
        visited[u] = true;
        int ans = INF;

        // Traverse through all
        // the adjacent nodes
        for (int i = 0; i < V; i++)
        {
            if (graph[u, i] != INF && !visited[i])
            {

                // Cost of the further path
                int curr = minimumCostSimplePath(i, destination,
                                                 visited, graph);

                // Check if we have reached the
                // destination
                if (curr < INF)
                {

                    // Taking the minimum cost path
                    ans = Math.Min(ans, graph[u, i] + curr);
                }
            }
        }

        // Unmarking the current node
        // to make it available for other
        // simple paths
        visited[u] = false;

        // Returning the minimum cost
        return ans;
    }

    // Driver code
    public static void Main(string[] args)
    {

        // Initialising the graph
        int[, ] graph = new int[V, V];
        for (int i = 0; i < V; i++)
        {
            for (int j = 0; j < V; j++)
            {
                graph[i, j] = INF;
            }
        }

        // Marking all nodes as unvisited
        bool[] visited = new bool[V];

        // Initialising the edges;
        graph[0, 1] = -1;
        graph[0, 3] = 1;
        graph[1, 2] = -2;
        graph[2, 0] = -3;
        graph[3, 2] = -1;
        graph[4, 3] = 2;

        // Source and destination
        int s = 0, t = 2;

        // Marking the source as visited
        visited[s] = true;
        Console.WriteLine(minimumCostSimplePath(s, t, visited, graph));
    }
}

// This code is contributed by sanjeev2552

Javascript

<script>

// JavaScript code for printing Minimum Cost
// Simple Path between two given nodes
// in a directed and weighted graph


// Define number of vertices in
// the graph and infinite value
let V = 5
let INF =  Number.MAX_SAFE_INTEGER

// Function to do DFS through the nodes
function minimumCostSimplePath(u, destination, visited, graph)
{

    // check if we find the destination
    // then further cost will be 0
    if (u == destination)
        return 0;

    // marking the current node as visited
    visited[u] = 1;

    let ans = INF;

    // traverse through all
    // the adjacent nodes
    for (let i = 0; i < V; i++) {
        if (graph[u][i] != INF && !visited[i]) {

            // cost of the further path
            let curr = minimumCostSimplePath(i,
                        destination, visited, graph);

            // check if we have reached the destination
            if (curr < INF) {

                // Taking the minimum cost path
                ans = Math.min(ans, graph[u][i] + curr);
            }
        }
    }

    // unmarking the current node
    // to make it available for other
    // simple paths
    visited[u] = 0;

    // returning the minimum cost
    return ans;
}

// driver code


    // initialising the graph
    let graph = new Array();
    for(let i = 0;  i< V; i++){
        graph.push([])
    }

    for (let i = 0; i < V; i++) {
        for (let j = 0; j < V; j++) {
            graph[i][j] = INF;
        }
    }

    // marking all nodes as unvisited
    let visited = new Array(V).fill(0);

    // initialising the edges;
    graph[0][1] = -1;
    graph[0][3] = 1;
    graph[1][2] = -2;
    graph[2][0] = -3;
    graph[3][2] = -1;
    graph[4][3] = 2;

    // source and destination
    let s = 0, t = 2;

    // marking the source as visited
    visited[s] = 1;

    document.write(minimumCostSimplePath(s, t, visited, graph));

    // This code is contributed by _saurabh_jaiswal

    </script>

Output:

-3

Time Complexity: O(V^2)
Auxiliary Space: O(V), since we are using an array of size V to store the visited nodes.