在给定图表中找到“足够短”的路径
我需要设计一种算法来寻找公共交通系统中的路径。从理论上讲，只需要最好的（最低成本）路径，但实际上它是不同的。在公共交通系统中旅行时，很难定义成本，它不能简化为旅行时间，等待时间，转移时间，公共汽车/地铁费等，都需要考虑。
首先，我需要简化问题，设计一个成本函数，它是所有那些“时间”和“费用”的组合，然后使用图算法找到几条路径（3~5路径）。最后向最终用户展示所有这些路径并让他们做出决定。
我需要提出多条路径的原因是，对于不同的用户/情况，这些“时间”和“费用”是不同的，因此提供一些路径比给出“最佳”路径更好。 / p>
像A *这样的算法很适合找到最短路径，但我怎样才能在图中找到那些“足够短”的路径？或者我怎样才能找到最短的N路径？
顺便说一下，我甚至不需要找到最短路径，因为在实践中最终用户永远不会知道最短路径（除非最短路径是明显的），如果结果接近最短路径，他们会很高兴2 个答案:
答案 0 :(得分：0)
A * star的“成本”比你想象的更通用。 A *通常用节点来解释，节点的成本只是距离。但是，我们可以加强一点。
我没有看到你喜欢的语言，也许是Graph？哦，这是一些c ++：
namespace Astar
{
struct CostEvaluation
{
int distance_cost;
int transfer_cost;
// others
int costToTraverseNodes( const Node& first, const Node& second ) const
{
int distance = // apply distance_cost to distance between nodes
int transfer = // apply transfer_cost if there is a transfer between nodes
return distance + transfer;
}
}
}
现在，A *的实际实施将采用CostEvaluation对象来确定路线上的成本。如果转移无关紧要，请将transfer_cost设置为零。
就“足够好”的路线而言：我相信其他人能够更好地帮助你，但我觉得你可能会遇到这样一种情况，即程序会说“哦，你想成为在一个小时内，但最好的路线只需要20分钟？在这里，绕圈四十分钟，这已经足够好了。“
答案 1 :(得分：0)
正如我在评论中暗示的那样，可以创建一个报告多条路线的修改后的A *版本。我只是把我的实现推到了显然确认这个陈述的程度。
下面的代码以“经典”A *实现开头，我保留了这个实现，因此可以研究“经典”和“修改”之间的差异。
修改版本的基本思想是并行开始向前和向后搜索。考虑到A *的“贪婪”主要由其启发式函数（h（x））驱动，这通常也会产生更强大的结果。有可能构建案例，其中贪婪选择在路线开始时快速进展，而朝向末端的路线“大幅减速”。从两侧（源，目标）开始，这种效果可以减轻到比特。 （如果计算到最后，它应该始终是最佳路线，如果不一定是相同的路线。如果要计算两个方向上的“经典”结束条件，可能会产生下面的图片，显示两者方向产生2条不同的路径。
现在，两个方向的“探索列表”可用于查找何时搜索，例如“前进”，下一个节点已经通过“向后”搜索进行了探索 - 反之亦然。显然，两次搜索之间的那些“连接点”产生了一条路线，这条路线不一定是最优的，而是一条有效的路线。
我的实现跟踪那些中间路线，我没有费心去收集它们。跟踪显示两个探索列表“相遇”的节点的id以及路由的结果2部分（源 - ＆gt;会合点，会合点 - ＆gt;目的地）。
现在，使用这些中间列表以及一些后期处理，例如通过根据启发函数的单个维度（例如舒适度，速度等）评估路线，应该可以找到足够好的路线选择，与这些维度中的不同权衡相关联。
完整的F＃脚本大约有340行 - 这个网站有点太长了，所以我省略了一些不必要的部分（例如我的渲染功能，创建这些位图等等。
module AStar =
module Internals =
let makeRoute (explo : Map<int,(int * float)>) at tgt =
let rec loop at acc =
let dst,c = explo.[at]
match at,dst with
| (_,b) when b = tgt -> (at,c) :: acc
| (_,b) -> loop b ((at,c) :: acc)
[(tgt,0.0)] @ loop at []
let makeRouteBackward (exploBW : Map<int, (int * float)>) at tgt =
let rec loop at acc =
let src,c = exploBW.[at]
match at,src with
| (_,b) when b = tgt -> acc @ [(at,c)]
| (_,b) -> loop b (acc @ [at,c])
let r = loop at [] @ [(tgt,0.0)]
let rev = List.rev r
List.zip r rev |> List.map (fun ((id1,c1),(id2,c2)) -> id1,c2)
let classic neighbors h cost start goal =
let prioSelect (lopen : (int * float) list) =
let sorted = List.sortBy (fun (id,p) -> p) lopen //|> List.rev
(fst (List.head sorted), List.tail sorted)
let rec search (lopen : (int * float) list) (routes : Map<int,int * float>) =
let rec searchNeighbors cur nl o (r : Map<int,(int * float)>) =
match nl with
| [] -> o,r
| next::others ->
let newCost = (snd (r.[cur])) + cost cur next
if (not (Map.containsKey next r)) || (newCost < snd r.[next])
then
let r1 = r |> Map.remove next |> Map.add next (cur,newCost)
let prio = newCost + h next goal
//printfn "current = %d -- next = %d -- newCost = %f -- prio = %f -- h = %f" cur next newCost prio (h next goal)
let o1 = (next,prio) :: o
searchNeighbors cur others o1 r1
else
searchNeighbors cur others o r
match lopen with
| [] -> []
| _::_ ->
let current,rest = prioSelect lopen
if current = goal then Internals.makeRoute routes current start
else
let lopen1,routes1 = searchNeighbors current (neighbors current) rest routes
search lopen1 routes1
search [start,0.] (Map.ofList [start,(start,0.0)])
let twoWay sources targets hforward hbackward costforward costbackward (start : int) (goal : int) (n : int) rr =
let prioSelect (lopen : (int * float) list) =
let sorted = List.sortBy (fun (id,p) -> p) lopen //|> List.rev
(fst (List.head sorted), List.tail sorted)
let searchforward lopen exploredF exploredB nfound acc =
let rec searchNeighbors cur nl o (r : Map<int,(int * float)>) =
match nl with
| [] -> o,r
| next::others ->
//printfn "fwd: current = %d -- next = %d -- nl = %A -- r = %A" cur next nl r
let newCost = (snd (r.[cur])) + costforward cur next
if (not (Map.containsKey next r)) || (newCost < snd r.[next])
then
let r1 = r |> Map.remove next |> Map.add next (cur,newCost)
let prio = newCost + hforward next goal
let o1 = (next,prio) :: o
if Map.containsKey next exploredB then
rr (next, Internals.makeRoute r1 next start, Internals.makeRouteBackward exploredB next goal)
searchNeighbors cur others o1 r1
else
searchNeighbors cur others o r
match lopen with
| [] -> (lopen,exploredF,0,acc)
| _::_ ->
let current,rest = prioSelect lopen
if current = goal then
(rest,exploredF,nfound+1,acc @ [Internals.makeRoute exploredF current start] )
else
let lopen1,explored1 = searchNeighbors current (targets current) rest exploredF
(lopen1, explored1, nfound, acc)
let searchbackward lopen exploredB exploredF nfound acc =
let rec searchNeighbors cur nl o (r : Map<int,(int * float)>) =
match nl with
| [] -> o,r
| next::others ->
//printfn "bwd: current = %d -- next = %d -- nl = %A -- r = %A" cur next nl r
let newCost = (snd (r.[cur])) + costbackward cur next
if (not (Map.containsKey next r)) || (newCost < snd r.[next])
then
let r1 = r |> Map.remove next |> Map.add next (cur,newCost)
let prio = newCost + hbackward next start
let o1 = (next,prio) :: o
searchNeighbors cur others o1 r1
else
searchNeighbors cur others o r
match lopen with
| [] -> (lopen,exploredB,0,acc)
| _::_ ->
let current,rest = prioSelect lopen
if current = start then
//(rest,explored,nfound+1,acc @ [Internals.makeRoute explored current goal []])
(rest,exploredB,nfound+1,acc @ [Internals.makeRouteBackward exploredB current goal] )
else
let lopen1,explored1 = searchNeighbors current (sources current) rest exploredB
(lopen1, explored1, nfound, acc)
let rec driver openF openB exploredF exploredB nfoundF nfoundB accF accB =
let openF1, exploredF1,nfoundF1,accF1 = searchforward openF exploredF exploredB nfoundF accF
let openB1, exploredB1,nfoundB1,accB1 = searchbackward openB exploredB exploredF nfoundB accB
match (nfoundF1+nfoundB1), List.isEmpty openF1, List.isEmpty openB1 with
| (s,false,false) when s < n ->
driver openF1 openB1 exploredF1 exploredB1 nfoundF1 nfoundB1 accF1 accB1
| _ ->
accF1 @ accB1
driver [start,0.0] [goal,0.0] (Map.ofList [start,(start,0.0)]) (Map.ofList [goal,(goal,0.0)]) 0 0 [] []
// Location : x,y coordinate or lat/long - whatever.
// Edges: (id,cost) list
type Node = { Id : int; Location : int * int; Edges : (int * float) list; EdgesBackward : (int * float) list}
type Graph = Map<int,Node>
let addNode node graph =
Map.add (node.Id) node graph
let newNode idgen x y =
{ Id = idgen(); Location = (x,y); Edges = []; EdgesBackward = [] }
let addEdge id cost node =
{ node with Node.Edges = node.Edges @ [(id,cost)]; }
let addEdgeBackward id cost node =
{ node with Node.EdgesBackward = node.EdgesBackward @ [(id,cost)]; }
let idgen startvalue =
let next = ref startvalue
fun () ->
let id = !next
next := !next + 1
id
let appendNode node nodeList = nodeList @ [node]
let sq x = x*x
let distance p1 p2 =
let x1,y1 = p1
let x2,y2 = p2
sqrt( float (sq (x2-x1) + sq (y2-y1)) )
let solve (g : Graph) s e =
let ns id =
g.[id].Edges |> List.map (fun (id,c) -> id)
let h at goal =
float (distance (g.[at].Location) (g.[goal].Location))
let c a b =
g.[a].Edges |> List.pick (fun (id,cost) -> if id = b then Some(cost) else None)
[AStar.classic ns h c s e] // give it the same return type as solveTwoWay to make stuff below easier and shorter
let solveTwoWay (g : Graph) s e n =
let edges id =
let nl = g.[id].Edges |> List.map (fun (id,c) -> id)
//printfn "2way edges id = %d list = %A" id nl
nl
let edgesBackward id =
let nl = g.[id].EdgesBackward |> List.map (fun (id,c) -> id)
//printfn "2way backwards edges id = %d list = %A" id nl
nl
let hforward at goal =
float (distance (g.[at].Location) (g.[goal].Location))
let hbackward at start =
float (distance (g.[at].Location) (g.[start].Location))
let costF a b =
g.[a].Edges |> List.pick (fun (id,cost) -> if id = b then Some(cost) else None)
let costB a b =
g.[a].EdgesBackward |> List.pick (fun (id,cost) -> if id = b then Some(cost) else None)
let debugView arg =
let id,r1,r2 = arg
printfn "meeting at %d: r1 = %A r2 = %A" id r1 r2
AStar.twoWay edgesBackward edges hforward hbackward costF costB s e n debugView
let solveProblem problem =
let g, start, goal = problem
g,start,goal,solve g start goal
let solveProblemTwoWay problem n =
let g, start, goal = problem
g,start,goal,solveTwoWay g start goal n
let save name solution =
let graph, start, goal, routes = solution
use writer = System.IO.File.CreateText("""E:\temp\""" + name + """.txt""")
fprintf writer "------------------------------------\n start = %d ----> goal = %d: %d routes found.\n" start goal (List.length routes)
fprintf writer "Graph:\n"
graph |> Map.iter
(fun id node ->
fprintf writer "Node: %A\n" node
)
routes |> List.iteri
(fun index route ->
fprintf writer "Route %d: %A\n" index route
)
// An example problem I used to play with:
// The graph is such, that the nodes are connected to the right and
// downwards and diagonally downwards only.
// The cost is either 1.0 or sqrt(2), for the horizontal or vertical and
// the diagonal connection, respectively.
let problem2 () =
let newNodeAN = newNode (idgen 0)
let cond c x n =
if c then n |> x else n
let accessCost p =
match p with
| (4,4) | (4,5) | (5,4) | (5,5) -> 10.0
| _ -> 1.0
let right (n : Node) : Node =
let t = 1 + fst n.Location, snd n.Location
let c = accessCost t
n
|> cond (fst n.Location < 9) (fun n -> addEdge (n.Id + 1) c n)
|> cond (fst n.Location > 0) (fun n -> addEdgeBackward (n.Id - 1) c n)
let down n =
let t = fst n.Location, 1 + snd n.Location
let c = accessCost t
n
|> cond (snd n.Location < 9) (fun n -> addEdge (n.Id + 10) c n)
|> cond (snd n.Location > 0) (fun n -> addEdgeBackward (n.Id - 10) c n)
let diagdown n =
let t = 1 + fst n.Location, 1 + snd n.Location
let c = (sqrt(2.0)) * accessCost t
n
|> cond (fst n.Location < 9 && snd n.Location < 9) (fun n -> addEdge (n.Id + 11) c n)
|> cond (fst n.Location > 0 && snd n.Location > 0) (fun n -> addEdgeBackward (n.Id - 11) c n)
[
for y = 0 to 9 do
for x = 0 to 9 do
yield newNodeAN x y
]
|> List.map
(fun n ->
n
|> right
|> down
|> diagdown
)
|> List.map (fun n -> (n.Id,n))
|> Map.ofList
, 0, 99
// Last not least, the code can be executed like this:
// And since both implementations yield the same data structures,
// they can be used interchangeably and compared to each other.
solveProblemTwoWay (problem2() 5) |> save "problem2_solution"
在运行时打印的输出显示“中间路径”，然后如下所示：
...
48日开会：
r1 = [（0,0.0）; （11,1.414213562）; （12,2.414213562）; （23,3.828427125）; （34,5.242640687）; （35,6.242640687）; （46,7.656854249）; （47,8.656854249）; （48,9.656854249）]
r2 = [（48,0.0）; （58,1.414213562）; （68,2.414213562）; （78,3.414213562）; （88,4.414213562）; （99,5.414213562）]
会议在84：
r1 = [（0,0.0）; （11,1.414213562）; （21,2.414213562）; （32,3.828427125）; （43,4.242640687）; （53,6.242640687）; （64,7.656854249）; （74,8.656854249）; （84,9.656854249）]
r2 = [（84,0.0）; （85,1.414213562）; （86,2.414213562）; （87,3.414213562）; （88,4.414213562）; （99,5.414213562）]
会议在95：
r1 = [（0,0.0）; （11,1.414213562）; （21,2.414213562）; （32,3.828427125）; （43,4.242640687）; （53,6.242640687）; （64,7.656854249）; （75,9.071067812）; （85,10.07106781）; （95,11.07106781）]
r2 = [（95,0.0）; （96,1.0）; （97,2.0）; （98,3.0）; （99,4.0）]
...
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