giles
3ea8967571
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Recursive 4-way flood fill from every unvisited 1-cell:
let rec flood visited r c =
if r < 0 || r >= h || c < 0 || c >= w then 0
else if visited.(r).(c) || grid.(r).(c) = 0 then 0
else begin
visited.(r).(c) <- true;
1 + flood visited (r - 1) c
+ flood visited (r + 1) c
+ flood visited r (c - 1)
+ flood visited r (c + 1)
end
Grid (1s shown as #, 0s as .):
# # . # #
# . . . #
. . # . .
# # # # .
. . . # #
Largest component: {(2,2),(3,0),(3,1),(3,2),(3,3),(4,3),(4,4)} = 7.
Bounds check r >= 0 must short-circuit before visited/grid reads;
relies on the && / || fix from iter 242.
157 baseline programs total.
39 lines
813 B
OCaml
39 lines
813 B
OCaml
let h = 5
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let w = 5
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let grid = [|
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[| 1; 1; 0; 1; 1 |];
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[| 1; 0; 0; 0; 1 |];
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[| 0; 0; 1; 0; 0 |];
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[| 1; 1; 1; 1; 0 |];
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[| 0; 0; 0; 1; 1 |]
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|]
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let rec flood visited r c =
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if r < 0 || r >= h || c < 0 || c >= w then 0
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else if visited.(r).(c) || grid.(r).(c) = 0 then 0
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else begin
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visited.(r).(c) <- true;
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1 + flood visited (r - 1) c
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+ flood visited (r + 1) c
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+ flood visited r (c - 1)
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+ flood visited r (c + 1)
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end
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let largest_component () =
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let visited = Array.init h (fun _ -> Array.make w false) in
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let best = ref 0 in
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for r = 0 to h - 1 do
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for c = 0 to w - 1 do
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if grid.(r).(c) = 1 && not visited.(r).(c) then begin
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let s = flood visited r c in
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if s > !best then best := s
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end
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done
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done;
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!best
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;;
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largest_component ()
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