# Copyright 2010-2018 Google LLC # Licensed under the Apache License, Version 2.0 (the "License"); # you may not use this file except in compliance with the License. # You may obtain a copy of the License at # # http://www.apache.org/licenses/LICENSE-2.0 # # Unless required by applicable law or agreed to in writing, software # distributed under the License is distributed on an "AS IS" BASIS, # WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. # See the License for the specific language governing permissions and # limitations under the License. # [START program] """From Taha 'Introduction to Operations Research', example 6.4-2.""" # [START import] from __future__ import print_function from ortools.graph import pywrapgraph # [END import] def main(): """MaxFlow simple interface example.""" # [START data] # Define three parallel arrays: start_nodes, end_nodes, and the capacities # between each pair. For instance, the arc from node 0 to node 1 has a # capacity of 20. start_nodes = [0, 0, 0, 1, 1, 2, 2, 3, 3] end_nodes = [1, 2, 3, 2, 4, 3, 4, 2, 4] capacities = [20, 30, 10, 40, 30, 10, 20, 5, 20] # [END data] # Instantiate a SimpleMaxFlow solver. # [START constraints] max_flow = pywrapgraph.SimpleMaxFlow() # Add each arc. for arc in zip(start_nodes, end_nodes, capacities): max_flow.AddArcWithCapacity(arc[0], arc[1], arc[2]) # [END constraints] # [START solve] # Find the maximum flow between node 0 and node 4. if max_flow.Solve(0, 4) == max_flow.OPTIMAL: print('Max flow:', max_flow.OptimalFlow()) print('') print(' Arc Flow / Capacity') for i in range(max_flow.NumArcs()): print('%1s -> %1s %3s / %3s' % (max_flow.Tail(i), max_flow.Head(i), max_flow.Flow(i), max_flow.Capacity(i))) print('Source side min-cut:', max_flow.GetSourceSideMinCut()) print('Sink side min-cut:', max_flow.GetSinkSideMinCut()) else: print('There was an issue with the max flow input.') # [END solve] if __name__ == '__main__': main() # [END program]