在C.B.的回答评论中，OP评论说：“请求是静态的。一开始我就得到了完整的列表。”如果我们提前获得所有旅行计划，我会欢迎反例和/或其他反馈，因为我认为如果我们考虑以下内容，问题可以大大减少：
由于电梯容量无限，任何向上的旅行终点低于我们将要访问的最高楼层的旅行都与我们的计算无关。由于我们保证经过所有这些接送点，我们可以在考虑下降旅行后将它们安排在我们的时间表中。
任何包含在同方向的其他旅行中的旅行也是无关紧要的，因为我们将在“最外层”旅行期间通过那些接送点，并且可以在考虑这些旅行后适当地安排它们。
任何重叠的下降旅行都可以合并，原因很快就会显现。
任何下降旅行都发生在达到最高点之前或之后（不包括达到最高楼层的接载）。我们确定发生在最高点之前的所有下降旅行（仅考虑“外部容器”类型和两个或更多重叠的旅行作为一个旅行）的最佳时间表是在我们上升时逐个进行，因为我们无论如何都要上升。
如何确定最高点后应该发生哪些下降行程？
我们将计算基准点 TOP 作为参考。我们称包含达到的最高楼层的行程为 H，最高楼层到达 HFR。如果 HFR 是上车点，则 H 是下降的，TOP = H_dropoff。如果 HFR 是下车点，则 H 是上升的，TOP = HFR。
应该在访问最高楼层之后安排的下降行程是从 TOP 下一个更低的下降行程开始向下继续，其中它们的组合距离加倍大于从 TOP 到它们最后一个下车点的总距离的相邻下降行程的最大组合（仅考虑“外部容器”类型和两个或多个重叠的行程作为一个行程）。也就是说，当 (D1 + D2 + D3...+ Dn) * 2 > TOP - Dn_dropoff 时。
以下是在 Haskell 中的粗略尝试：
import Data.List (sort,sortBy)
trips = [(101,100),(50,49),(25,19),(99,97),(95,93),(30,20),(35,70),(28,25)]
isDescending (a,a') = a > a'
areDescending a b = isDescending a && isDescending b
isContained aa@(a,a') bb@(b,b') = areDescending aa bb && a < b && a' > b'
extends aa@(a,a') bb@(b,b') = areDescending aa bb && a <= b && a > b' && a' < b'
max' aa@(a,a') bb@(b,b') = if (maximum [b,a,a'] == b) || (maximum [b',a,a'] == b')
then bb
else aa
(outerDescents,innerDescents,ascents,topTrip) = foldr f ([],[],[],(0,0)) trips where
f trip (outerDescents,innerDescents,ascents,topTrip) = g outerDescents trip ([],innerDescents,ascents,topTrip) where
g [] trip (outerDescents,innerDescents,ascents,topTrip) = (trip:outerDescents,innerDescents,ascents,max' trip topTrip)
g (descent:descents) trip (outerDescents,innerDescents,ascents,topTrip)
| not (isDescending trip) = (outerDescents ++ (descent:descents),innerDescents,trip:ascents,max' trip topTrip)
| isContained trip descent = (outerDescents ++ (descent:descents),trip:innerDescents,ascents,topTrip)
| isContained descent trip = (trip:outerDescents ++ descents,descent:innerDescents,ascents,max' trip topTrip)
| extends trip descent = ((d,t'):outerDescents ++ descents,(t,d'):innerDescents,ascents,max' topTrip (d,t'))
| extends descent trip = ((t,d'):outerDescents ++ descents,(d,t'):innerDescents,ascents,max' topTrip (t,d'))
| otherwise = g descents trip (descent:outerDescents,innerDescents,ascents,topTrip)
where (t,t') = trip
(d,d') = descent
top = snd topTrip
scheduleFirst descents = (sum $ map (\(from,to) -> 2 * (from - to)) descents)
> top - (snd . last) descents
(descentsScheduledFirst,descentsScheduledAfterTop) =
(descentsScheduledFirst,descentsScheduledAfterTop) where
descentsScheduledAfterTop = (\x -> if not (null x) then head x else [])
. take 1 . filter scheduleFirst
$ foldl (\accum num -> take num sorted : accum) [] [1..length sorted]
sorted = sortBy(\a b -> compare b a) outerDescents
descentsScheduledFirst = if null descentsScheduledAfterTop
then sorted
else drop (length descentsScheduledAfterTop) sorted
scheduled = ((>>= \(a,b) -> [a,b]) $ sort descentsScheduledFirst)
++ (if isDescending topTrip then [] else [top])
++ ((>>= \(a,b) -> [a,b]) $ sortBy (\a b -> compare b a) descentsScheduledAfterTop)
place _ [] _ _ = error "topTrip was not calculated."
place floor' (floor:floors) prev (accum,numStops)
| floor' == prev || floor' == floor = (accum ++ [prev] ++ (floor:floors),numStops)
| prev == floor = place floor' floors floor (accum,numStops)
| prev < floor = f
| prev > floor = g
where f | floor' > prev && floor' < floor = (accum ++ [prev] ++ (floor':floor:floors),numStops)
| otherwise = place floor' floors floor (accum ++ [prev],numStops + 1)
g | floor' < prev && floor' > floor = (accum ++ [prev] ++ (floor':floor:floors),numStops)
| otherwise = place floor' floors floor (accum ++ [prev],numStops + 1)
schedule trip@(from,to) floors = take num floors' ++ fst placeTo
where placeFrom@(floors',num) = place from floors 1 ([],1)
trimmed = drop num floors'
placeTo = place to (tail trimmed) (head trimmed) ([],1)
solution = foldl (\trips trip -> schedule trip trips) scheduled (innerDescents ++ ascents)
main = do print trips
print solution
输出：
*Main> main
[(101,100),(50,49),(25,19),(99,97),(95,93),(30,20),(35,70),(28,25)]
[1,25,28,30,25,20,19,35,50,49,70,101,100,99,97,95,93]
