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Yanqing from Lipson Community College and Andrei from Tudor Vianu National College, Bucharest, Romania both sent correct solutions.
Andrei wrote:
" As I am not familiar with graph theory, I first read the hint and the notes of the problem, then I looked on the web for the Maximum flow, minimum cut theorem. I found different examples, algorithms of solving such problems, as well a possible list of utilisation.
I started with the definition of terms: a cut in a network is a minimal set of edges whose removal separates the network into two components, one containing the source, and the other the sink."
In the first network a cut through PT, RT, RU, RB, and SB separates the network showing that the flow cannot be more than 18 units (the total flow through these edges).
This flow is possible by sending: 3 units along ASB, 5 units along ASRB, 5 units along ARB, 2 units along ARUTB, 1 unit along AQRTB and 2 units along APTB. Hence the maximum flow is 18 units.
In the second network a cut through BC, DC, FG and HG separates the network so the flow cannot be more than 16 units (the total flow along these edges).
This flow is possible by sending 3 units along ABFG, 4 units along ABCG, 2 units along ADCG, 3 units along ADHG and 4 units along AEHG. Hence the maximum flow is 16 units.
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