Answers for "Djikstra's algorithm"

9

java djikstra's algorithm

import java.util.*; 
public class DPQ { 
    private int dist[]; 
    private Set<Integer> settled; 
    private PriorityQueue<Node> pq; 
    private int V; // Number of vertices 
    List<List<Node> > adj; 
  
    public DPQ(int V) 
    { 
        this.V = V; 
        dist = new int[V]; 
        settled = new HashSet<Integer>(); 
        pq = new PriorityQueue<Node>(V, new Node()); 
    } 
  
    // Function for Dijkstra's Algorithm 
    public void dijkstra(List<List<Node> > adj, int src) 
    { 
        this.adj = adj; 
  
        for (int i = 0; i < V; i++) 
            dist[i] = Integer.MAX_VALUE; 
  
        // Add source node to the priority queue 
        pq.add(new Node(src, 0)); 
  
        // Distance to the source is 0 
        dist[src] = 0; 
        while (settled.size() != V) { 
  
            // remove the minimum distance node  
            // from the priority queue  
            int u = pq.remove().node; 
  
            // adding the node whose distance is 
            // finalized 
            settled.add(u); 
  
            e_Neighbours(u); 
        } 
    } 
  
    // Function to process all the neighbours  
    // of the passed node 
    private void e_Neighbours(int u) 
    { 
        int edgeDistance = -1; 
        int newDistance = -1; 
  
        // All the neighbors of v 
        for (int i = 0; i < adj.get(u).size(); i++) { 
            Node v = adj.get(u).get(i); 
  
            // If current node hasn't already been processed 
            if (!settled.contains(v.node)) { 
                edgeDistance = v.cost; 
                newDistance = dist[u] + edgeDistance; 
  
                // If new distance is cheaper in cost 
                if (newDistance < dist[v.node]) 
                    dist[v.node] = newDistance; 
  
                // Add the current node to the queue 
                pq.add(new Node(v.node, dist[v.node])); 
            } 
        } 
    } 
  
    // Driver code 
    public static void main(String arg[]) 
    { 
        int V = 5; 
        int source = 0; 
  
        // Adjacency list representation of the  
        // connected edges 
        List<List<Node> > adj = new ArrayList<List<Node> >(); 
  
        // Initialize list for every node 
        for (int i = 0; i < V; i++) { 
            List<Node> item = new ArrayList<Node>(); 
            adj.add(item); 
        } 
  
        // Inputs for the DPQ graph 
        adj.get(0).add(new Node(1, 9)); 
        adj.get(0).add(new Node(2, 6)); 
        adj.get(0).add(new Node(3, 5)); 
        adj.get(0).add(new Node(4, 3)); 
  
        adj.get(2).add(new Node(1, 2)); 
        adj.get(2).add(new Node(3, 4)); 
  
        // Calculate the single source shortest path 
        DPQ dpq = new DPQ(V); 
        dpq.dijkstra(adj, source); 
  
        // Print the shortest path to all the nodes 
        // from the source node 
        System.out.println("The shorted path from node :"); 
        for (int i = 0; i < dpq.dist.length; i++) 
            System.out.println(source + " to " + i + " is "
                               + dpq.dist[i]); 
    } 
} 
  
// Class to represent a node in the graph 
class Node implements Comparator<Node> { 
    public int node; 
    public int cost; 
  
    public Node() 
    { 
    } 
  
    public Node(int node, int cost) 
    { 
        this.node = node; 
        this.cost = cost; 
    } 
  
    @Override
    public int compare(Node node1, Node node2) 
    { 
        if (node1.cost < node2.cost) 
            return -1; 
        if (node1.cost > node2.cost) 
            return 1; 
        return 0; 
    } 
}
Posted by: Guest on March-27-2020
4

dijkstra algorithm c++

#include<bits/stdc++.h>
using namespace std;

int main()
{
	int n = 9;
	
	int mat[9][9] = { { 100,4,100,100,100,100,100,8,100}, 
                      { 4,100,8,100,100,100,100,11,100}, 
                      {100,8,100,7,100,4,100,100,2}, 
                      {100,100,7,100,9,14,100,100,100}, 
                      {100,100,100,9,100,100,100,100,100}, 
                      {100,100,4,14,10,100,2,100,100}, 
                      {100,100,100,100,100,2,100,1,6}, 
                      {8,11,100,100,100,100,1,100,7}, 
                      {100,100,2,100,100,100,6,7,100}};
	
	int src = 0;
	int count = 1;
	
	int path[n];
	for(int i=0;i<n;i++)
		path[i] = mat[src][i];
	
	int visited[n] = {0};
	visited[src] = 1;
	
	while(count<n)
	{
		int minNode;
		int minVal = 100;
		
		for(int i=0;i<n;i++)
			if(visited[i] == 0 && path[i]<minVal)
			{
				minVal = path[i];
				minNode = i;
			}
		
		visited[minNode] = 1;
		
		for(int i=0;i<n;i++)
			if(visited[i] == 0)
				path[i] = min(path[i],minVal+mat[minNode][i]);
					
		count++;
	}
	
	path[src] = 0;
	for(int i=0;i<n;i++)
		cout<<src<<" -> "<<path[i]<<endl;
	
	return(0);
}
Posted by: Guest on August-29-2020
0

dijkstra's algorithm

function Dijkstra(Graph, source):
       dist[source]  := 0                     // Distance from source to source is set to 0
       for each vertex v in Graph:            // Initializations
           if v ≠ source
               dist[v]  := infinity           // Unknown distance function from source to each node set to infinity
           add v to Q                         // All nodes initially in Q

      while Q is not empty:                  // The main loop
          v := vertex in Q with min dist[v]  // In the first run-through, this vertex is the source node
          remove v from Q 

          for each neighbor u of v:           // where neighbor u has not yet been removed from Q.
              alt := dist[v] + length(v, u)
              if alt < dist[u]:               // A shorter path to u has been found
                  dist[u]  := alt            // Update distance of u 

      return dist[]
  end function
Posted by: Guest on August-12-2021
1

dijkstra's algorithm

# Providing the graph
n = int(input("Enter the number of vertices of the graph"))

# using adjacency matrix representation 
vertices = [[0, 0, 1, 1, 0, 0, 0],
            [0, 0, 1, 0, 0, 1, 0],
            [1, 1, 0, 1, 1, 0, 0],
            [1, 0, 1, 0, 0, 0, 1],
            [0, 0, 1, 0, 0, 1, 0],
            [0, 1, 0, 0, 1, 0, 1],
            [0, 0, 0, 1, 0, 1, 0]]

edges = [[0, 0, 1, 2, 0, 0, 0],
         [0, 0, 2, 0, 0, 3, 0],
         [1, 2, 0, 1, 3, 0, 0],
         [2, 0, 1, 0, 0, 0, 1],
         [0, 0, 3, 0, 0, 2, 0],
         [0, 3, 0, 0, 2, 0, 1],
         [0, 0, 0, 1, 0, 1, 0]]

# Find which vertex is to be visited next
def to_be_visited():
    global visited_and_distance
    v = -10
    for index in range(num_of_vertices):
        if visited_and_distance[index][0] == 0 \
            and (v < 0 or visited_and_distance[index][1] <=
                 visited_and_distance[v][1]):
            v = index
    return v


num_of_vertices = len(vertices[0])

visited_and_distance = [[0, 0]]
for i in range(num_of_vertices-1):
    visited_and_distance.append([0, sys.maxsize])

for vertex in range(num_of_vertices):

    # Find next vertex to be visited
    to_visit = to_be_visited()
    for neighbor_index in range(num_of_vertices):

        # Updating new distances
        if vertices[to_visit][neighbor_index] == 1 and 
                visited_and_distance[neighbor_index][0] == 0:
            new_distance = visited_and_distance[to_visit][1] 
                + edges[to_visit][neighbor_index]
            if visited_and_distance[neighbor_index][1] > new_distance:
                visited_and_distance[neighbor_index][1] = new_distance
        
        visited_and_distance[to_visit][0] = 1

i = 0

# Printing the distance
for distance in visited_and_distance:
    print("Distance of ", chr(ord('a') + i),
          " from source vertex: ", distance[1])
    i = i + 1
Posted by: Guest on April-17-2021

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