what dijkstra's algo does code example

Example 1: dijkstra algorithm

//djikstra's algorithm using a weighted graph (STL)
//code by Soumyadepp
//insta: @soumyadepp
//linkedinID: https://www.linkedin.com/in/soumyadeep-ghosh-90a1951b6/

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

//to find the closest unvisited vertex from the source
//note that numbering of vertices starts from 1 here. Calculate accordingly
ll minDist(ll dist[], ll n, bool visited[])
{
    ll min = INT_MAX;
    ll minIndex = 0;
    for (ll i = 1; i <= n; i++)
    {
        if (!visited[i] && dist[i] <= min)
        {
            min = dist[i];
            minIndex = i;
        }
    }
    return minIndex;
}

//djikstra's algorithm for single source shortest path
void djikstra(vector<pair<ll, ll>> *g, ll n, ll src)
{
    bool visited[n + 1];
    ll dist[n + 1];
    for (ll i = 0; i <= n; i++)
    {
        dist[i] = INT_MAX;
        visited[i] = false;
    }

    dist[src] = 0;

    for (ll i = 0; i < n - 1; i++)
    {
        ll u = minDist(dist, n, visited);
        visited[u] = true;
        for (ll v = 0; v < g[u].size(); v++)
        {
            if (dist[u] + g[u][v].second < dist[g[u][v].first])
            {
                dist[g[u][v].first] = dist[u] + g[u][v].second;
            }
        }
    }
    cout << "VERTEX : DISTANCE" << endl;
    for (ll i = 1; i <= n; i++)
    {
        if (dist[i] != INT_MAX)
            cout << i << "         " << dist[i] << endl;
        else
            cout << i << "         "
                 << "not reachable" << endl;
    }
    cout << endl;
}

int main()
{
    //to store the adjacency list which also contains the weight
    vector<pair<ll, ll>> *graph;
    ll n, e, x, y, w, src;
    cout << "Enter number of vertices and edges in the graph" << endl;
    cin >> n >> e;
    graph = new vector<pair<ll, ll>>[n + 1];
    cout << "Enter edges and weight" << endl;
    for (ll i = 0; i < e; i++)
    {
        cin >> x >> y >> w;
        //checking for invalid edges and negative weights.
        if (x <= 0 || y <= 0 || w <= 0)
        {
            cout << "Invalid parameters. Exiting" << endl;
            exit(-1);
        }
        graph[x].push_back(make_pair(y, w));
        graph[y].push_back(make_pair(x, w));
    }
    cout << "Enter source from which you want to find shortest paths" << endl;
    cin >> src;
    if (src >= 1 && src <= n)
        djikstra(graph, n, src);
    else
        cout << "Please enter a valid vertex as the source" << endl;
    return 0;
}

//time complexity : O(ElogV)
//space complexity: O(V)

Example 2: 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