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author | Carlo Zancanaro <carlo@carlo-laptop> | 2012-11-09 12:17:46 +1100 |
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committer | Carlo Zancanaro <carlo@carlo-laptop> | 2012-11-09 12:17:46 +1100 |
commit | b19bd8d8a41664328f33c9b459b2b0100f0b303f (patch) | |
tree | c4ef4e35fe92d357cb75bbb38dadcddd52e26dde /clang/test/Lexer/counter.c | |
parent | 1e907d883191571b5c374fd1c3b2f6c1fe11da83 (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.
Diffstat (limited to 'clang/test/Lexer/counter.c')
0 files changed, 0 insertions, 0 deletions