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我从codereview交叉发布这个问题，因为我发现它没有响应。
此问题在hackerrank ai中可用我不是在寻求解决方案，而是试图找出我的策略或代码有什么问题。
我正试图解决一个我认为是TSP on a 2-D grid的问题。所以，我在尽力取得最好的成绩。然而，向前看1步比向前看2步产生更好的结果。
问题是我必须以最少的移动次数来清理二维网格上的脏块。
另一个重要的事情是，我移动了一步，然后处理的是UP, DOWN, LEFT, RIGHT, CLEAN与网格的新状态和我的新位置。所以我必须重新运行算法。这也意味着我必须避免陷入循环，这在单进程的情况下很容易避免，但在进程的多个实例的情况下需要通过算法来保证。
简言之，在我的过程中，我只需要做restarted。
所以基本的策略是找到离我现在位置最近的肮脏牢房。
向前看一步，我会做：为每个脏细胞，并找到最近的脏细胞采取脏细胞对于2步，对于每个脏单元格，执行1步查找并找到最佳移动。同样，对于多个步骤。
但是，当我只执行一步查找，而执行两步查找时，我得到的分数会更高。分数由next_move计算。所以，我认为我的代码/策略有问题。
输入格式：(200 - steps_taken)表示网格中的bot。b是干净的细胞-是脏电池。
第一行是机器人位置的点对这使得网格中的d冗余如果bot当前位于脏单元上，则网格中的该单元上会出现b。
第二条线是网格的维数。
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代码是：
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
我控制如何使用变量
d。样板是：
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
deep和deep调用新进程，直到我清除所有脏单元。我做错什么了？
最佳答案
经过大量的工作，我找到了一个最佳路径确定的例子
在级别1上的byfindBestMove返回的路径比使用
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只探索了一个可能的最佳动作。我发现如果你把所有的
最好的动作如下：
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 ]
这里
greedy结合了findCost和playGame只看一眼排序列表中的第一步
playGame取决于排序算法以及点的顺序。
你也可以这样写：
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 ) ]
这更清楚地强调了两种情况之间的对称性。
下面是我用来查找和显示上述板的最佳路径的代码：
http://lpaste.net/121294要使用它，只需将代码放入名为
bestMove的模块中。最后，我的直觉告诉我，你的做法可能不是一个好办法
来解决这个问题。你所做的与a*-算法相似
其中，
Ashish起着启发作用。然而，为了使a*起作用，启发式函数不应高估
最短的距离。但是
playGame总是给你一个上界最短的距离无论如何，这是值得考虑的事情。
08-24 13:46