2 Given a n*n adjacency array. it will give you a maximum flow. This version use BFS to search path. 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 19 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]]
2 Given a n*n adjacency array. it will give you a maximum flow. This version use BFS to search path. 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 19 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]]
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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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 - 0(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
while True:
#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
while bfs.qsize() > 0:
#pop from queue
src = bfs.get()
for k in range (len(new_array)):
#checking capacity and visit
if(new_array[src] [k]> 0 and visited[k] == 0 ):
#if not, put into queue and chage to visit and save path
visited[k] = 1
bfs.put(k)
path[k] = src](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3437af43-6a1b-4faa-a426-da58e71127ca%2F11eb9e6b-901d-4f0f-869d-93a536c7b8c9%2Fr2l6o1v_processed.png&w=3840&q=75)
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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 - 0(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
while True:
#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
while bfs.qsize() > 0:
#pop from queue
src = bfs.get()
for k in range (len(new_array)):
#checking capacity and visit
if(new_array[src] [k]> 0 and visited[k] == 0 ):
#if not, put into queue and chage to visit and save path
visited[k] = 1
bfs.put(k)
path[k] = src
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