Solution Review: Implementing Dijkstra's

def Dijkstra(graph, src, dst):
    number_of_nodes = len(graph[0])   # Number of nodes in the graph
    parent = [-1] * number_of_nodes    # Setting up various lists
    visited = []                    
    unvisited = [i for i in range(number_of_nodes)] 
    distance = [16] * number_of_nodes # The distance list initially has a distance of infinite for all but the source node
    distance[src] = 0
    shortest_path = []                # This list will have the shortest path at the end
    current = src                     # We start with the source
    while(len(unvisited)>0):   
      # Visit all neighbors of current and update distance
      for i in range(number_of_nodes):
        if(graph[current][i]>=0 and distance[i] > graph[current][i]+distance[current]):
          distance[i] = graph[current][i]+distance[current] # Update distance
          parent[i] = current # Set new parent
      unvisited.remove(current) # Move current node from unvisited to visted
      visited.append(current)
      if(len(unvisited) != 0):
        current = unvisited[0] # New current node is an unvisited node with the smallest 'distance'
        for n in unvisited:
          if(distance[n] < distance[current]):
            current = n
    curr = dst # Some code to get the shortest path from 'parent' 
    shortest_path.append(curr)
    cost = 0
    while curr is not src:
      if parent[curr] == -1: # If there is no path to the source node
        return([[],-1])
      cost = cost + graph[curr][parent[curr]] # The cost is the sum of the links in a path
      curr = parent[curr]
      shortest_path.append(curr)
    shortest_path.reverse()
    return([shortest_path, cost])
def main():
    graph = [
      [0,1,5,-1,-1],
      [1,0,3,-1,9],
      [5,3,0,4,-1],
      [-1,-1,4,0,2],
      [-1,9,-1,2,0]
    ]
    src = 0
    dst = 3
    print(Dijkstra(graph,src,dst))
if __name__ == "__main__":
    main()