Program to find total cost for completing all shipments in python

PythonServer Side ProgrammingProgramming

Suppose we have a list of lists called ports, where ports[i] represents the list of ports that port i is connected to. We also have another list of lists called shipments where each list of the sequence [i, j] which denotes there is a shipment request from port i to port j. And the cost to ship from port i to port j is the length of the shortest path from the two ports, we have to find the total cost necessary to complete all the shipments.

So, if the input is like ports = [[1, 4],[2],[3],[0, 1],[]] shipments = [[1, 4]], then the output will be 4, as the path is from 1 -> 2 -> 3 -> 0 -> 4.

To solve this, we will follow these steps - 

  • n := size of ports
  • dist := adjacency matrix from the ports list
  • for j in range 0 to n, do
    • for i in range 0 to n, do
      • for k in range 0 to n, do
        • dist[i, k] = minimum of dist[i, k], dist[i, j] + dist[j, k]
  • for all shipments in form [i, j] make a list dist[i,j] when dist[i,j] is not infinity
  • return the sum of generated list.

Let us see the following implementation to get better understanding:


Live Demo

class Solution:
   def solve(self, ports, shipments):
      n = len(ports)
      INF = 10 ** 10
      dist = [[INF for _ in range(n)] for _ in range(n)]
      for i in range(n):
         dist[i][i] = 0
      for i in range(n):
         for j in ports[i]:
            dist[i][j] = 1
      for j in range(n):
         for i in range(n):
            for k in range(n):
               dist[i][k] = min(dist[i][k], dist[i][j] + dist[j][k])

      return sum(dist[i][j] for i, j in shipments if dist[i][j] != INF)

ob = Solution()
ports = [[1, 4],[2],[3],[0, 1],[]]
shipments = [[1, 4]]
print(ob.solve(ports, shipments))


[[1, 4],[2],[3],[0, 1],[]], [[1, 4]]


Published on 02-Dec-2020 09:16:54