t will give you a maximum flow. This version use BFS to search path. Assume the first is the source and the last is the sink. Time complexity - O(Ef) example graph = [[0, 16, 13, 0, 0, 0], [0, 0, 10, 12, 0, 0], [0, 4, 0, 0, 14, 0],

Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Given a n*n adjacency array.
  it will give you a maximum flow.
  This version use BFS to search path.
   
  Assume the first is the source and the last is the sink.
   
  Time complexity - O(Ef)
   
  example
   
  graph = [[0, 16, 13, 0, 0, 0],
  [0, 0, 10, 12, 0, 0],
  [0, 4, 0, 0, 14, 0],
  [0, 0, 9, 0, 0, 20],
  [0, 0, 0, 7, 0, 4],
  [0, 0, 0, 0, 0, 0]]
   
  answer should be
   
  23
   
  """
  import copy
  import queue
  import math
   
  def maximum_flow_bfs(adjacency_matrix):
  """
  Get the maximum flow through a graph using a breadth first search
  """
  #initial setting
  new_array=copy.deepcopy(adjacency_matrix)
  total=0
   
  whileTrue:
  #setting min to max_value
  min_flow=math.inf
  #save visited nodes
  visited= [0]*len(new_array)
  #save parent nodes
  path= [0]*len(new_array)
   
  #initialize queue for BFS
  bfs=queue.Queue()
   
  #initial setting
  visited[0] =1
  bfs.put(0)
   
  #BFS to find path
  whilebfs.qsize() >0:
  #pop from queue
  src=bfs.get()
  forkinrange(len(new_array)):
  #checking capacity and visit
  if(new_array[src][k] >0andvisited[k] ==0 ):
  #if not, put into queue and chage to visit and save path
  visited[k] =1
  bfs.put(k)
  path[k] =src
   
  #if there is no path from src to sink
  ifvisited[len(new_array) -1] ==0:
  break
   
  #initial setting
 

tmp=len(new_array) -1. 

 

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