Haskell，204271字节
编辑：通过将凸包更新为列表（但与非高尔夫版本具有相同的复杂性），使用“ split（x + 1）”而不是“ splitLookup x”，并删除所有像Predule这样的合格函数调用，可以进一步打高尔夫球。折叠
这将创建一个函数f，该函数期望一个（a，b）对的列表和一个x值的列表。我想，使用相同的想法，APL系列中的任何东西都将在长度方向上将其吹灭，但实际上是这样：
import Data.Map
r=fromListWith max
[]%v=[(0,v)]
i@((p,u):j)%v|p>v#u=j%v|0<1=(v#u,v):i
(a,b)#(c,d)=1+div(b-d)(c-a)
o i x=(\(a,b)->a*x+b)$snd$findMax$fst$split(x+1)$r$foldl'(%)[]$r$zip(fmap fst i)i
f=fmap.o
用法示例：
[1 of 1] Compiling Main ( linear-min.hs, interpreted )
Ok, modules loaded: Main.
λ> f [(2,8), (4,0), (2,1), (1,10), (3,3), (0,4)] [1..5]
[11,12,14,16,20]
λ> f [(1,20), (2,12), (3,11), (4,8)] [1..5]
[21,22,23,24,28]
它的工作时间为O（n log n）；请参阅下面的分析。
编辑：这是一个带有big-O分析的简单版本，并对其全部工作方式进行了描述：
import Prelude hiding (null, empty)
import Data.Map hiding (map, foldl)
-- Just for clarity:
type X = Int
type Y = Int
type Line = (Int,Int)
type Hull = Data.Map.Map X Line
slope (a,b) = a
{-- Take a list of pairs (a,b) representing lines a*x + b and sort by
ascending slope, dropping any lines which are parallel to but below
another line.
This composition is O(n log n); n for traversing the input and
the output, log n per item for dictionary inserts during construction.
The input and output are both lists of length <= n.
--}
sort :: [Line] -> [Line]
sort = toList . fromListWith max
{-- For lines ax+b, a'x+b' with a < a', find the first value of x
at which a'x + b' exceeds ax + b. --}
breakEven :: Line -> Line -> X
breakEven p@(a,b) q@(a',b') = if slope p < slope q
then 1 + ((b - b') `div` (a' - a))
else error "unexpected ordering"
{-- Update the convex hull with a line whose slope is greater
than any other lines in the hull. Drop line segments
from the hull until the new line intersects the final segment.
split is used to find the portion of the convex hull
to the right of some x value; it has complexity O(log n).
insert is also O(log n), so one invocation of this
function has an O(log n) cost.
updateHull is recursive, but see analysis for hull to
account for all updateHull calls during one execution.
--}
updateHull :: Line -> Hull -> Hull
updateHull line hull
| null hull = singleton 0 line
| slope line <= slope lastLine = error "Hull must be updated with lines of increasing slope"
| hull == hull' = insert x line hull
| otherwise = updateHull line hull''
where (lastBkpt, lastLine) = findMax hull
x = breakEven lastLine line
hull' = fst $ x `split` hull
hull'' = fst $ lastBkpt `split` hull
{-- Build the convex hull by adding lines one at a time,
ordered by increasing slope.
Each call to updateHull has an immediate cost of O(log n),
and either adds or removes a segment from the hull. No
segment is added more than once, so the total cost is
O(n log n).
--}
hull :: [Line] -> Hull
hull = foldl (flip updateHull) empty . sort
{-- Find the highest line for the given x value by looking up the nearest
(breakpoint, line) pair with breakpoint <= x. This uses the neat
function splitLookup which looks up a key k in a dictionary and returns
a triple of:
- The subdictionary with keys < k.
- Just v if (k -> v) is in the dictionary, or Nothing otherwise
- The subdictionary with keys > k.
O(log n) for dictionary lookup.
--}
valueOn :: Hull -> X -> Y
valueOn boundary x = a*x + b
where (a,b) = case splitLookup x boundary of
(_ , Just ab, _) -> ab
(lhs, _, _) -> snd $ findMax lhs
{-- Solve the problem!
O(n log n) since it maps an O(log n) function over a list of size O(n).
Computation of the function to map is also O(n log n) due to the use
of hull.
--}
solve :: [Line] -> [X] -> [Y]
solve lines = map (valueOn $ hull lines)
-- Test case from the original problem
test = [(2,8), (4,0), (2,1), (1,10), (3,3), (0,4)] :: [Line]
verify = solve test [1..5] == [11,12,14,16,20]
-- Test case from comment
test' = [(1,20),(2,12),(3,11),(4,8)] :: [Line]
verify' = solve test' [1..5] == [21,22,23,24,28]