Camino con el menor producto de aristas con peso >= 1

Dado un grafo dirigido con N Nodes y E aristas donde el peso de cada arista es > 1 , también dado un origen S y un destino D . La tarea es encontrar el camino con el mínimo producto de aristas de S a D. Si no hay una ruta de S a D , imprima -1 .

Ejemplos:  

Entrada: N = 3, E = 3, Bordes = {{{1, 2}, 5}, {{1, 3}, 9}, {{2, 3}, 1}}, S = 1 y D = 3 
Salida: 5 
El camino con el menor producto de aristas será 1->2->3 
con un producto de 5*1 = 5.

Entrada: N = 3, E = 3, Bordes = {{{3, 2}, 5}, {{3, 3}, 9}, {{3, 3}, 1}}, S = 1 y D = 3 
Salida: -1 

Enfoque: La idea es utilizar el algoritmo de ruta más corta de Dijkstra con una ligera variación. 
Se pueden seguir los siguientes pasos para calcular el resultado:  

• Si el origen es igual al destino, devuelve 0 .
• Inicialice una cola de prioridad pq ​​con S y su peso como 1 y una array visitada v[] .
• Mientras que pq no está vacío: 
  1. Haga estallar el elemento superior de pq . Llamémoslo como curr y su producto de distancia como dist .
  2. Si ya se ha visitado curr , continúe.
  3. Si curr es igual a D , devuelve dist .
  4. Iterar todos los Nodes adyacentes a curr y presionar en pq (siguiente y dist + gr[nxt].weight)
• devolver -1 .

A continuación se muestra la implementación del enfoque anterior:  

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the smallest
// product of edges
double 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 code
int 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;
}

// Java implementation of the approach
import 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 edges
static 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 code
public 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

# 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

<script>
 
// Javascript implementation of the approach
 
// Function to return the smallest
// product of edges
function dijkstra(s, d, gr)
{
    // If the source is equal
    // to the destination
    if (s == d)
        return 0;
 
    // Initialise the priority queue
    var pq = [];
    pq.push([1, s]);
 
    // Visited array
    var v = Array(gr.length).fill(0);
 
    // While the priority-queue
    // is not empty
    while (pq.length!=0) {
 
        // Current node
        var curr = pq[0][1];
 
        // Current product of distance
        var dist = pq[0][0];
 
        // Popping the top-most element
        pq.shift();
 
        // 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 (var it of gr[curr])
            pq.push([dist * it[1], it[0] ]);
        pq.sort();
    }
 
    // If no path exists
    return -1;
}
 
// Driver code
var n = 3;
 
// Graph as adjacency matrix
var gr = Array.from(Array(n + 1), ()=> Array());
 
// Input edges
gr[1].push([3, 9]);
gr[2].push([3, 1]);
gr[1].push([2, 5]);
 
// Source and destination
var s = 1, d = 3;
 
// Dijkstra
document.write(dijkstra(s, d, gr));
 
// This code is contributed by rrrtnx.
</script>

5

Complejidad temporal: O((E + V) logV)
Espacio auxiliar : O(V).