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authorCarlo Zancanaro <carlo@carlo-laptop>2012-11-09 12:17:46 +1100
committerCarlo Zancanaro <carlo@carlo-laptop>2012-11-09 12:17:46 +1100
commitb19bd8d8a41664328f33c9b459b2b0100f0b303f (patch)
treec4ef4e35fe92d357cb75bbb38dadcddd52e26dde /clang/test/SemaObjC/conflict-nonfragile-abi2.m
parent1e907d883191571b5c374fd1c3b2f6c1fe11da83 (diff)
Add an MCF operator to the separate solver
For the solver utility it'd be good to have MCF problems, so here they are! Format is: MCF<supplies, arcs>(cost*) Supplies is a [int,int,int,...], where each int represents a new node Arcs is [int:int, int:int, int:int, ...] where each int:int pair represents an edge from the first to the second (1 indexed from the "supplies" array). Costs is the argument to the function. There must be as many costs as arcs, and they are set from left to right, in order.
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