我正在codereview交叉发布此问题，因为我发现它没有响应。
此问题可在hackerrank ai处获得。我不是要求解决方案，而是试图找出我的策略或代码有什么问题。
我正在尝试解决一个我认为是TSP on a 2-D grid的问题。所以，我正在努力获得最好的结果。然而，展望未来的一步是产生比前两步更好的结果。
问题是我必须以最小的移动次数清除2-D网格上的脏块UP, DOWN, LEFT, RIGHT, CLEAN。
另一件重要的事情是，我采取行动，然后处理restarted新的网格状态和我的新位置。所以我必须再次运行算法。这也意味着我必须避免处于循环中，这在单个进程的情况下很容易避免，但是在多个进程实例的情况下需要由算法保证。
简而言之，我必须在我的流程中只制作next_move。
所以基本策略是找到离我当前位置最近的脏单元格。
要向前查看1步，我会这样做：对于每个脏单元，找到最接近脏单元的脏单元。对于每个脏单元格的2步，执行一步查找并找到最佳移动。同样适用于多个步骤。
然而，当我只进行一步查找但是两步查找得分较少时，我得分较高。分数由(200 - steps_taken)计算。所以，我认为我的代码/策略有问题。
输入格式：
b代表网格中的机器人。 -是干净的细胞。 d是脏单元格。
第一行是机器人位置的一对。这使得网格b变得冗余。如果bot当前站在脏单元格上，则d将出现在网格中的该单元格上。
第二行是网格的维度。
第三个输入是行格式的网格。请参阅下面的示例输入。
我的Haskell代码：
module Main where
import Data.List
import Data.Function (on)
import Data.Ord
-- slits up a string
-- ** only used in IO.
split sep = takeWhile (not . null) . unfoldr (Just . span (/= sep) . dropWhile (== sep))
-- ** only used in IO
getList :: Int -> IO [String]
getList n = if n==0 then return [] else do i <- getLine; is <- getList(n-1); return (i:is)
-- find positions of all dirty cells in the board
getAllDirtyCells :: (Int, Int) -> [String] -> [(Int, Int)]
getAllDirtyCells (h, w) board = [(x, y) | x <- [0..(h-1)], y <- [0..(w - 1)]
, ((board !! x) !! y) == 'd']
-- finally get the direction to print ;
-- first argument is my-position and second arg is next-position.
getDir :: (Int, Int) -> (Int, Int) -> String
getDir (x, y) (a, b) | a == x && y == b = "CLEAN"
| a < x = "UP"
| x == a && y < b = "RIGHT"
| x == a = "LEFT"
| otherwise = "DOWN"
-- only used in IO for converting strin gto coordinate.
getPos :: String -> (Int, Int)
getPos pos = let a = map (\x -> read x :: Int) (words pos)
in ((a !! 0) , (a !! 1))
-- manhattan Distance : sum of difference of x and y coordinates
manhattanDis :: (Int, Int) -> (Int, Int) -> Int
manhattanDis (a, b) (x, y) = (abs (a - x) + (abs (b - y)))
-- sort the positions from (botX, botY) position on manhattan-distance.
-- does not returns the cost.
getSortedPos :: (Int, Int) -> [(Int, Int)] -> [(Int, Int)]
getSortedPos (botX, botY) points = map (\x -> (snd x)) $
sortBy (comparing fst) -- compare on the basis of cost.
[(cost, (a, b)) |
(a, b) <- points,
cost <- [manhattanDis (a,b) (botX, botY)]]
-- exclude the point `v` from the list `p`
excludePoint :: (Ord a) => [a] -> a -> [a]
excludePoint [] _ = []
excludePoint p v = [x | x <- p , x /= v]
-- playGame uses the nearest-node-policy.
-- we start playing game when we are not going more deep.
-- more about that in findBestMove
-- game is to reduce the nodes to one node with the total cost ;
-- reduction : take the next shortest node from the current-node.
playGame :: (Int, Int) -> [(Int, Int)] -> [(Int, Int)]
playGame pos [] = [pos]
playGame startPos points = let nextPos = (head (getSortedPos startPos points))
in (nextPos : playGame nextPos (excludePoint points nextPos))
-- sum up cost of all the points as they occur.
findCost :: [(Int, Int)] -> Int
findCost seq = sum $ map (\x -> (manhattanDis (fst x) (snd x))) $ zip seq (tail seq)
-- find the position which gives the smallest overall cost.
smallestCostMove :: [(Int, (Int, Int))] -> (Int, (Int, Int))
smallestCostMove [] = (0, (100, 100))
smallestCostMove [x] = x
smallestCostMove (x:y:xs) | (fst x) <= (fst y) = smallestCostMove (x : xs)
| otherwise = smallestCostMove (y : xs)
-- This is actual move-finder. It does the lookups upto `level` deep.
-- from startpoint, take each point and think it as starting pos and play the game with it.
-- this helps us in looking up one step.
-- when level is 0, just use basic `playGame` strategy.
findBestMove :: (Int, Int) -> [(Int, Int)] -> Int -> (Int, (Int, Int))
findBestMove startPos points level
-- returns the move that takes the smallest cost i.e. total distances.
| level == 0 = smallestCostMove $
-- return pair of (cost-with-node-x-playGame, x)
map (\x -> (findCost (startPos : (x : (playGame x (excludePoint points x)))),
x))
points
| otherwise = smallestCostMove $
map (\x ->
-- return pair of (cost-with-node-x, x)
( (findCost (startPos : [x])) +
-- findBestMove returns the pair of (cost, next-move-from-x)
(fst (findBestMove x (excludePoint points x) (level - 1))),
x))
points
-- next_move is our entry point. go only 2 level deep for now, as it can be time-expensive.
next_move :: (Int, Int) -> (Int, Int) -> [String] -> String
next_move pos dim board = let boardPoints = (getAllDirtyCells dim board)
numPoints = (length boardPoints)
-- ** Important : This is my question :
-- change the below `deep` to 1 for better results.
deep = if (numPoints > 3)
then 2
else if (numPoints == 1)
then 1
else (numPoints - 1)
in if pos `elem` boardPoints
then getDir pos pos
else getDir pos $ snd $ findBestMove pos boardPoints deep
main :: IO()
main = do
-- Take input
b <- getLine
i <- getLine
-- bot contains (Int, Int) : my-coordinates. like (0,0)
let botPos = (read $ head s::Int,read $ head $ tail s::Int) where s = split (' ') b
-- dimOfBoard contains dimension of board like (5,5)
let dimOfBoard = (read $ head s::Int, read $ head $ tail s::Int) where s = split (' ') i
board <- getList (fst dimOfBoard)
putStrLn $ next_move botPos dimOfBoard board
我控制deep我可以使用变量deep的方式。
示例板是：
0 0
5 5
b---d
-d--d
--dd-
--d--
----d
可以有三个答案：
输出：
RIGHT or DOWN or LEFT
重要：
再次使用new board和my bot new position调用新进程，直到我清理所有脏单元格。
我做错了什么？
答案 0 :(得分：1)
经过大量工作后，我找到了确定最佳路径的例子
在{1}处的findBestMove返回的路径比调用时更糟
级别设置为0：
points = [(6,8),(9,7),(9,4),(4,10),(4,6),(7,10),(5,7),(2,4),(8,8),(6,5)]
start: (1,10)
level = 0:
cost: 31
path: [(1,10),(4,10),(7,10),(5,7),(6,8),(8,8),(9,7),(9,4),(6,5),(4,6),(2,4)]
level = 1:
cost: 34
path: [(1,10),(2,4),(6,5),(6,8),(5,7),(4,6),(4,10),(7,10),(8,8),(9,7),(9,4)]
问题是playGame只探讨了最好的一个动作。
我发现如果你探索所有的算法，你的算法会更稳定
最好的举动如下：
greedy start [] = 0
greedy start points =
let sorted@((d0,_):_) = sort [ (dist start x, x) | x <- points ]
nexts = map snd $ takeWhile (\(d,_) -> d == d0) sorted
in d0 + minimum [ greedy n (delete n points) | n <- nexts ]
此处greedy结合了findCost和playGame。只看着
排序列表playGame中的第一个移动取决于排序算法
以及点的排序。
您也可以这样写bestMove：
bestMove _ start [] = (0,start)
bestMove depth start points
| depth == 0 = minimum [ (d0+d,x) | x <- points,
let d0 = dist start x,
let d = greedy x (delete x points) ]
| otherwise = minimum [ (d0+d,x) | x <- points,
let d0 = dist start x,
let (d,_) = bestMove (depth-1) x (delete x points ) ]
这更清楚地突出了两种情况之间的对称性。
以下是我用来查找并显示上述主板最佳路径的代码：
http://lpaste.net/121294要使用它，只需将代码放在名为Ashish的模块中。
最后，我的直觉告诉我，你的方法可能不是一个好方法
解决问题。你在做什么类似于A *算法
playGame扮演启发式函数的角色。然而，
为了使A *起作用，启发式函数不应过度估计
最短的距离。但playGame总是给你一个上限
最短的距离。无论如何 - 这是值得考虑的事情。