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 nodesint 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 codeint 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 graphimport java.util.*;import java.lang.*;class GFG{ // Define number of vertices in// the graph and infinite valuestatic int V = 5;static int INF = Integer.MAX_VALUE;// Function to do DFS through the nodesstatic 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 codepublic 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 graphimport sysV = 5INF = sys.maxsize # Function to do DFS through the nodesdef 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 codeif __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 graphusing 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 valuelet V = 5let INF = Number.MAX_SAFE_INTEGER// Function to do DFS through the nodesfunction 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.
Please Login to comment...