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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:
-3
Time Complexity: O(V^2)
Auxiliary Space: O(V), since we are using an array of size V to store the visited nodes.