Maximum weighted bipartite matching _with_ directed edges

南楼画角 提交于 2019-12-06 11:11:24

dfb's comment is correct, for any two vertices A, B you can discard the cheaper of the two edges AB and BA.

The proof is a one-liner:

Theorem: A maximum matching M never contains the cheaper edge of AB and BA for any two vertices A,B.

Proof: Let M be a maximum matching. Suppose AB is in M and is cheaper than BA. Define M' = M - {AB} + {BA}. M' is clearly still a matching, but it's more expensive. That contraditcs the assumption that M was a maximum matching.

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