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