One directed graph is provided with the weight between each pair of vertices, and two vertices u and v are also provided. Our task is to find the shortest distance from vertex u to vertex v, with exactly k number of edges.
To solve this problem, we will start from vertex u and go to all adjacent vertices and recur for adjacent vertices using the k value as k - 1.
Input and Output
Input: The cost matrix of the graph. 0 10 3 2 ∞ 0 ∞ 7 ∞ ∞ 0 6 ∞ ∞ ∞ 0 Output: Weight of the shortest path is 9
Algorithm
shortKEdgePath(u, v, edge)
Input − Vertex u and v, and a number of edges.
Output − Distance of shortest path.
Begin if edge = 0 and u = v, then return 0 if edge = 1 and cost[u, v] ≠ ∞, then return cost[u, v] if edge <= 0, then return ∞ set shortPath := ∞ for all vertices i, do if cost[u, i] ≠ ∞ and u ≠ i and v ≠ i, then tempRes := shortKEdgePath(i, v, edge - 1) if tempRes ≠ ∞, then shortPath = minimum of shortPath and (cost[u,i]+tempRes done return shortPath End
Example
#include <iostream> #define NODE 4 #define INF INT_MAX using namespace std; int cost[NODE][NODE] = { {0, 10, 3, 2}, {INF, 0, INF, 7}, {INF, INF, 0, 6}, {INF, INF, INF, 0} }; int minimum(int a, int b) { return (a<b)?a:b; } int shortKEdgePath(int u, int v, int edge) { if (edge == 0 && u == v) //when 0 edge, no path is present return 0; if (edge == 1 && cost[u][v] != INF) //when only one edge, and (u,v) is valid return cost[u][v]; if (edge <= 0) //when edge is -ve, there are infinity solution return INF; int shortPath = INF; for (int i = 0; i < NODE; i++) { //for all vertices i, adjacent to u if (cost[u][i] != INF && u != i && v != i) { int tempRes = shortKEdgePath(i, v, edge-1); if (tempRes != INF) shortPath = minimum(shortPath, cost[u][i] + tempRes); } } return shortPath; } int main() { int src = 0, dest = 3, k = 2; cout << "Weight of the shortest path is " << shortKEdgePath(src, dest, k); }
Output
Weight of the shortest path is 9