GraphHopper如何处理高速公路交叉口上的点?

本文介绍了GraphHopper如何处理高速公路交叉口上的点?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我在比利时有2750个市中心.我需要知道每两个城市中心之间的距离. 但这导致了57MB的矩阵，只是为了记住这些距离(甚至不包括路线)，因此可以进行大规模扩展.

I have 2750 city centers in Belgium. I need to know the distances between every 2 city centers. But that results in an matrix of 57MB, just to remember those distances (not even the routes), so that scales terribly.

相反，我正在考虑使用公路交叉口作为枢纽.基本上，每个城市都知道附近的城市和附近的枢纽(=公路交叉点).所有集线器都知道彼此的距离.

Instead, I am looking at using Highway intersections as hubs. Basically, every city knows it's nearby cities and it's nearby hubs (= highway intersection). All hubs know the distance to each other.

因此，可以通过cityA -> hubX -> hubY -> cityB的距离来计算从一个城市A到另一个非邻近城市B的距离.因为大多数城市通常在附近有3个枢纽，所以我可能需要查看所有9种组合并以最短的组合.但无论如何，它应该明智地扩展内存.

So the distance from 1 city A to another non-nearby city B, can be calculated by the distance of cityA -> hubX -> hubY -> cityB. Because most cities have typically 3 hubs nearby, I might need to look at all 9 combinations and take the shortest. But in any case it should scale better memory wise.

现在的问题是: 我可以将高速公路交叉口描述为一个点吗?考虑一下:高速公路由2条道路组成(双向)，所以高速公路交叉点中心有4条道路(甚至不算武器)

Now the problem: Can I describe a highway intersection as a single point? Think about it: a highway consist of 2 roads (one in both direction), so a highway intersection center has 4 roads (not even counting the arms).

推荐答案

一些想法:

您可以通过MapDB或磁盘将这些距离存储在堆外或盘上 GraphHopper其简单的DataAccess实现使其与RAM无关

您可以使用浮点数，它应该只有〜30MB甚至很短，仅用公里数即可.

您可以尝试按需路由，而无需存储，因为它只需几毫秒即可计算出一条路由.禁用指令，计算点数使其速度提高两倍.您甚至可以禁用计算距离，而仅使用path.weight-这将为您带来另一个良好的加速效果，但需要使用较低级别的GraphHopper，并且仅在知道您的操作后才建议使用.

you can store those distances off-heap or on-disc via MapDB or GraphHopper its simplistic DataAccess implementations making it RAM-independent

you can use float which should be only ~30MB or even short and just use the kilometers

you could try on demand routing, without storing, as it takes only a few ms to calculate a route. Disable instructions and calculating points makes it even twice as fast. You could even disable calculating the distance and just use path.weight - this will give you another good speedup but requires a bit lower level GraphHopper usage and is only recommended if you know what you do.

现在是您的问题. GraphHopper使用图模型，该图模型由节点(结)和边(连接结的街道)组成.环形交叉路口仍由多个节点组成.但总的来说，应该可以将离开"节点用作"hub-id".

Now to your question. GraphHopper uses a graph model consisting of nodes (junctions) and edges (streets connecting junctions). Still a roundabout consists of multiple nodes. But in general it should be possible to use such a 'leaving' node as 'hub-id'.

我看到了两种计算这些节点的方法:

I see two approaches to calculate those nodes:

• 通过运行收缩层次结构"并选择最高的1000个节点并将它们定义为集线器，这类似于在运输节点路由'论文
• 或者您计算从一个城市到例如所有其他城市(或仅8个地理方向)并找到两条路线的最后一个公共节点以识别一些

• either by running the Contraction-Hierarchy and picking the highest 1000 nodes and define them as hubs - this would be similar to what is described in the 'transit node routing' paper
• or you calculate routes from one city to e.g. all other cities (or just 8 geographic directions) and find the last common nodes of two routes to identify some

对于这两种方法，您都必须对GraphHopper进行更深入的研究，并且可能需要较低级别的API .

For both approaches you'll have to digg a bit deeper into GraphHopper and you'll probably need the lower level API.