Path with smallest product of edges with weight >= 1
Given a directed graph with N nodes and E edges where the weight of each of the edge is > 1, also given a source S and a destination D. The task is to find the path with the minimum product of edges from S to D. If there is no path from S to D then print -1.
Examples:
Input: N = 3, E = 3, Edges = {{{1, 2}, 5}, {{1, 3}, 9}, {{2, 3}, 1}}, S = 1, and D = 3
Output: 5
The path with smallest product of edges will be 1->2->3
with product as 5*1 = 5.Input: N = 3, E = 3, Edges = {{{3, 2}, 5}, {{3, 3}, 9}, {{3, 3}, 1}}, S = 1, and D = 3
Output: -1
Approach: The idea is to use Dijkstra’s shortest path algorithm with a slight variation.
The following steps can be followed to compute the result:
- If the source is equal to the destination then return 0.
- Initialise a priority-queue pq with S and its weight as 1 and a visited array v[].
- While pq is not empty:
- Pop the top-most element from pq. Let’s call it as curr and its product of distance as dist.
- If curr is already visited then continue.
- If curr is equal to D then return dist.
- Iterate all the nodes adjacent to curr and push into pq (next and dist + gr[nxt].weight)
- return -1.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the smallest// product of edgesdouble dijkstra(int s, int d, vector<vector<pair<int, double> > > gr){ // If the source is equal // to the destination if (s == d) return 0; // Initialise the priority queue set<pair<int, int> > pq; pq.insert({ 1, s }); // Visited array bool v[gr.size()] = { 0 }; // While the priority-queue // is not empty while (pq.size()) { // Current node int curr = pq.begin()->second; // Current product of distance int dist = pq.begin()->first; // Popping the top-most element pq.erase(pq.begin()); // If already visited continue if (v[curr]) continue; // Marking the node as visited v[curr] = 1; // If it is a destination node if (curr == d) return dist; // Traversing the current node for (auto it : gr[curr]) pq.insert({ dist * it.second, it.first }); } // If no path exists return -1;}// Driver codeint main(){ int n = 3; // Graph as adjacency matrix vector<vector<pair<int, double> > > gr(n + 1); // Input edges gr[1].push_back({ 3, 9 }); gr[2].push_back({ 3, 1 }); gr[1].push_back({ 2, 5 }); // Source and destination int s = 1, d = 3; // Dijkstra cout << dijkstra(s, d, gr); return 0;} |
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Java
// Java implementation of the approachimport java.util.ArrayList;import java.util.PriorityQueue;class Pair implements Comparable<Pair> { int first, second; public Pair(int first, int second) { this.first = first; this.second = second; } public int compareTo(Pair o) { if (this.first == o.first) { return this.second - o.second; } return this.first - o.first; }}class GFG{// Function to return the smallest// xor sum of edgesstatic int dijkstra(int s, int d, ArrayList<ArrayList<Pair>> gr){ // If the source is equal // to the destination if (s == d) return 0; // Initialise the priority queue PriorityQueue<Pair> pq = new PriorityQueue<>(); pq.add(new Pair(1, s)); // Visited array boolean[] v = new boolean[gr.size()]; // While the priority-queue // is not empty while (!pq.isEmpty()) { // Current node Pair p = pq.poll(); int curr = p.second; // Current xor sum of distance int dist = p.first; // If already visited continue if (v[curr]) continue; // Marking the node as visited v[curr] = true; // If it is a destination node if (curr == d) return dist; // Traversing the current node for(Pair it : gr.get(curr)) pq.add(new Pair(dist ^ it.second, it.first)); } // If no path exists return -1;}// Driver codepublic static void main(String[] args){ int n = 3; // Graph as adjacency matrix ArrayList<ArrayList<Pair>> gr = new ArrayList<>(); for(int i = 0; i < n + 1; i++) { gr.add(new ArrayList<Pair>()); } // Input edges gr.get(1).add(new Pair(3, 9)); gr.get(2).add(new Pair(3, 1)); gr.get(1).add(new Pair(2, 5)); // Source and destination int s = 1, d = 3; System.out.println(dijkstra(s, d, gr));}}// This code is contributed by sanjeev2552 |
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Python3
# Python3 implementation of the approach # Function to return the smallest # product of edges def dijkstra(s, d, gr) : # If the source is equal # to the destination if (s == d) : return 0; # Initialise the priority queue pq = []; pq.append(( 1, s )); # Visited array v = [0]*(len(gr) + 1); # While the priority-queue # is not empty while (len(pq) != 0) : # Current node curr = pq[0][1]; # Current product of distance dist = pq[0][0]; # Popping the top-most element pq.pop(); # If already visited continue if (v[curr]) : continue; # Marking the node as visited v[curr] = 1; # If it is a destination node if (curr == d) : return dist; # Traversing the current node for it in gr[curr] : if it not in pq : pq.insert(0,( dist * it[1], it[0] )); # If no path exists return -1; # Driver code if __name__ == "__main__" : n = 3; # Graph as adjacency matrix gr = {}; # Input edges gr[1] = [( 3, 9) ]; gr[2] = [ (3, 1) ]; gr[1].append(( 2, 5 )); gr[3] = []; #print(gr); # Source and destination s = 1; d = 3; # Dijkstra print(dijkstra(s, d, gr)); # This code is contributed by AnkitRai01 |
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Output:
5
Time complexity: O((E + V) logV)
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