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Haskell Range Minimum Query
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import Data.List.Split | |
import qualified Data.Vector as V | |
rmqBuild :: [Int] -> (Int, V.Vector Int) | |
rmqBuild xs = | |
let l = length xs | |
n = until (\ x -> x >= l) (*2) 1 | |
inf = maxBound :: Int | |
f ys = map minimum $ chunksOf 2 ys | |
in (n ,V.fromList $ concat $ | |
until (\x -> (length (head x)) == 1) | |
(\(y:ys) -> ((f y):y:ys)) ([xs ++ take (n - l) [inf, inf..]])) | |
rmqQuery n tree a b = rmqQuery' tree a b 0 0 n | |
where | |
rmqQuery' tree a b k l r = | |
if r <= a || b <= l | |
then maxBound | |
else if a <= l && r <= b | |
then tree V.! k | |
else min (rmqQuery' tree a b (2 * k + 1) l ((l + r) `div` 2)) | |
(rmqQuery' tree a b (2 * k + 2) ((l + r) `div` 2) r) | |
main = do | |
contents <- getContents | |
let ds = lines contents | |
(n, t) = rmqBuild (map (read :: String -> Int) (words (ds !! 1))) | |
qs = map (\x -> map (read :: String -> Int) (words x)) (drop 2 ds) | |
mapM_ (\x -> putStrLn $ show (rmqQuery n t (x !! 0) ((x !! 1) + 1) :: Int)) | |
qs | |
{-| test input | |
10 5 | |
10 20 30 40 11 22 33 44 15 5 | |
0 5 | |
1 2 | |
8 9 | |
0 9 | |
4 6 | |
-} |
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