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对于我的学习，我必须编写以下函数，它可以获得两个国家之间的最短路径。我已经写过一个函数isRoute来检查两个国家之间是否存在连接，而函数yieldRoute只是返回两个国家之间的连接。 现在我必须编写一个返回两个国家之间最短路线的函数。如何在Haskell中实现Dijkstra算法
我的第一个方法是获得两个国家之间的所有联系，然后获得最短的联系，但在我看来，获得所有联系对程序员来说是一件烦人的事情。现在我想出实现dijstra算法的想法，但实际上我也觉得这很难。你们可以给我一些想法如何做到这一点？
我们必须使用这些类型
type Country = String
type Countries = [Country]
type TravelTime = Integer -- Travel time in minutes
data Connection = Air Country Country TravelTime
| Sea Country Country TravelTime
| Rail Country Country TravelTime
| Road Country Country TravelTime deriving (Eq,Ord,Show)
type Connections = [Connection]
data Itinerary = NoRoute | Route (Connections,TravelTime) deriving (Eq,Ord,Show)
我屈服路由功能这简直是广度优先搜索（我们不能改变他们，但我们OFC允许添加新类型）：（Sry基因德国评论）
-- Liefert eine Route falls es eine gibt
yieldRoute :: Connections -> Country -> Country -> Connections
yieldRoute cons start goal
| isRoute cons start goal == False = []
| otherwise = getRoute cons start [] [start] goal
getRoute :: Connections -> Country -> Connections -> Countries -> Country -> Connections
getRoute cons c gone visited target
| (c == target) = gone
| otherwise = if (visit cons c visited) then (getRoute cons (deeper cons c visited) (gone ++ get_conn cons c (deeper cons c visited)) (visited ++ [(deeper cons c visited)]) target) else (getRoute cons (back (drop (length gone -1) gone)) (take (length gone -1) gone) visited target)
-- Geht ein Land zurück
back :: Connections -> Country
back ((Air c1 c2 _):xs) = c1
back ((Sea c1 c2 _):xs) = c1
back ((Rail c1 c2 _):xs) = c1
back ((Road c1 c2 _):xs) = c1
-- Liefert das nächste erreichbare Country
deeper :: Connections -> Country -> Countries -> Country
deeper ((Air c1 c2 _):xs) c visited
| (c1 == c) = if (c2 `elem` visited) then (deeper xs c visited) else c2
| (c2 == c) = if (c1 `elem` visited) then (deeper xs c visited) else c1
| otherwise = deeper xs c visited
deeper ((Sea c1 c2 _):xs) c visited
| (c1 == c) = if (c2 `elem` visited) then (deeper xs c visited) else c2
| (c2 == c) = if (c1 `elem` visited) then (deeper xs c visited) else c1
| otherwise = deeper xs c visited
deeper ((Rail c1 c2 _):xs) c visited
| (c1 == c) = if (c2 `elem` visited) then (deeper xs c visited) else c2
| (c2 == c) = if (c1 `elem` visited) then (deeper xs c visited) else c1
| otherwise = deeper xs c visited
deeper ((Road c1 c2 _):xs) c visited
| (c1 == c) = if (c2 `elem` visited) then (deeper xs c visited) else c2
| (c2 == c) = if (c1 `elem` visited) then (deeper xs c visited) else c1
| otherwise = deeper xs c visited
-- Liefert eine Connection zwischen zwei Countries
get_conn :: Connections -> Country -> Country -> Connections
get_conn [] _ _ = error "Something went terribly wrong"
get_conn ((Air c1 c2 t):xs) c3 c4
| (c1 == c3) && (c2 == c4) = [(Air c1 c2 t)]
| (c1 == c4) && (c2 == c3) = [(Air c1 c2 t)]
| otherwise = get_conn xs c3 c4
get_conn ((Sea c1 c2 t):xs) c3 c4
| (c1 == c3) && (c2 == c4) = [(Air c1 c2 t)]
| (c1 == c4) && (c2 == c3) = [(Air c1 c2 t)]
| otherwise = get_conn xs c3 c4
get_conn ((Road c1 c2 t):xs) c3 c4
| (c1 == c3) && (c2 == c4) = [(Air c1 c2 t)]
| (c1 == c4) && (c2 == c3) = [(Air c1 c2 t)]
| otherwise = get_conn xs c3 c4
get_conn ((Rail c1 c2 t):xs) c3 c4
| (c1 == c3) && (c2 == c4) = [(Air c1 c2 t)]
| (c1 == c4) && (c2 == c3) = [(Air c1 c2 t)]
| otherwise = get_conn xs c3 c4
-- Überprüft ob eine besuchbare Connection exestiert
visit :: Connections -> Country -> Countries -> Bool
visit [] _ _ = False
visit ((Air c1 c2 _):xs) c visited
| (c1 == c) = if (c2 `elem` visited) then (visit xs c visited) else True
| (c2 == c) = if (c1 `elem` visited) then (visit xs c visited) else True
| otherwise = visit xs c visited
visit ((Sea c1 c2 _):xs) c visited
| (c1 == c) = if (c2 `elem` visited) then (visit xs c visited) else True
| (c2 == c) = if (c1 `elem` visited) then (visit xs c visited) else True
| otherwise = visit xs c visited
visit ((Rail c1 c2 _):xs) c visited
| (c1 == c) = if (c2 `elem` visited) then (visit xs c visited) else True
| (c2 == c) = if (c1 `elem` visited) then (visit xs c visited) else True
| otherwise = visit xs c visited
visit ((Road c1 c2 _):xs) c visited
| (c1 == c) = if (c2 `elem` visited) then (visit xs c visited) else True
| (c2 == c) = if (c1 `elem` visited) then (visit xs c visited) else True
这一次我现在写的：
yieldFastestRoute :: Connections -> Country -> Country -> Itinerary
Dijkst RA算法： http://en.wikipedia.org/wiki/Dijkstra%27s_algorithm
我的第一种方法是这样的：（正如我所说的，我与getallRoutes问题）
yieldFastestRoute :: Connections -> Country -> Country -> Itinerary
yieldFastestRoute cons start targ
|(isRoute start targ == False) = NoRoute
|otherwise = (Route (getFastest (getAllRoutes cons start targ)) (sumTT (getFastest (getAllRoutes cons start targ))))
-- Liefert alle Routen zwischen zwei Ländern
getAllRoutes :: Connections -> Country -> Country -> [Connections]
-- Liefert aus einer Reihe von Connections die schnellste zurück
getFastest :: [Connections] -> Connections
getFastest (x:xs) = if ((sumTT x) < sumTT (getFastest xs) || null (getFastest xs)) then x else (getFastest xs)
sumTT :: Connections -> TravelTime
sumTT [] = 0
sumTT ((Air _ _ t): xs) = t ++ sumTT xs
sumTT ((Rail _ _ t): xs) = t ++ sumTT xs
sumTT ((Road _ _ t): xs) = t ++ sumTT xs
sumTT ((Sea _ _ t): xs) = t ++ sumTT xs
我基本上想知道什么是最好的方式在Haskell实现Dijkstra算法，或者如果还有另一种方法可以遵循。
1.什么是Dijkstra算法？向我们展示你实施它的尝试。 3.说明实施它的哪一部分你觉得很困难。 – dave4420
我想如果theres在haskell中实现dijstra并不是一个极端困难的方式，或者如果有一些更容易的方法来解决问题： http://en.wikipedia.org/wiki/Dijkstra%27s_algorithm –
我想这个问题会如果你专注于如何创建适当的图形数据结构，那么应该更好地回答。在那之后，实施Dijkstra应该不难。此外，您还有大量的代码，并且有点难以吞咽，特别是德语注释 – hugomg