Haskell, 204 271 Bytes
Bearbeiten : Golf viel weiter durch Aktualisieren der konvexen Hülle als Liste (aber mit der gleichen Komplexität wie die ungolfed Version), Verwenden von "split (x + 1)" anstelle von "splitLookup x" und Entfernen aller qualifizierten Funktionsaufrufe wie Predule. foldl.
Dies erzeugt eine Funktion f, die die Liste von (a, b) Paaren und eine Liste von x Werten erwartet. Ich denke, es wird von irgendetwas in der APL-Familie mit den gleichen Ideen in die Länge getrieben, aber hier ist es:
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
Beispielnutzung:
[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]
Es funktioniert in O (n log n) Zeit; Analyse siehe unten.
Bearbeiten: Hier ist eine ungolfed Version mit der Big-O-Analyse und einer Beschreibung, wie alles funktioniert:
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]