ICPC by Haskell

ながながとICPC2005 Regional, Tokyo の問題Fを Haskell で．O(n^2) の DP で, 基本的に foldl の２重ネストと．Cで書いた場合と比べて for 文の代わりに foldl になっているだけの違いしかない...

 -- Problem F in ACM/ICPC 2005 ASIA Regional Tokyo
 -- 2005/11/04  O(n^2) DP
import Control.Monad
import Debug.Trace
import List
main = getProblems >>= mapM_ (putStrLn.show.solve)
 
getProblems = 
    do
    [n] <- getNums
    if n==0 then return []
       else do
            xs <- getNums
            b <- liftM (head.map read.words) getLine :: IO(Double)
            [sr,sv,se,sf] <- liftM words getLine
            liftM ((xs, b, read sr, read sv::Double, read se::Double, read sf::Double):) getProblems
    where 
    getNums = liftM (map read.words) getLine :: IO([Int])
 
solve :: ([Int], Double, Int, Double, Double, Double) -> Double
solve (ys,b,r,v,e,f) = sl ys [(0.0,0)] 
    where
    -- xs:rest of checkpoints, cs:best cost for already passed checkpoints 
    sl [] cs = fst $ head cs
    sl (x:xs) cs = sl xs (best x cs:cs)
    -- (best cost, current cost, current point)
    best x cs = (\(bc,c,p) -> (min bc (c + cost (x-p) x),x)) $ foldl op (100000000, 0, x) cs
      where 
      op (cbc, c, p) (bc, pos) = let
                                 c' = c + cost (x-p) (x-pos)
                                 in (min cbc (c'+b+bc), c', pos)
    cost p q = ct p 0.0
      where 
      ct p c = if p == q then c
               else ct (p+1) (c+step p)
      step p = if p < r then 1.0 / (v-f*fromIntegral(r-p))
               else  1.0 / (v-e*fromIntegral(p-r))