Probably similar to the answer on pre-computed routes between major locations and layered maps, but my understanding is that in games, to speed up A*, you have a map that is very coarse for macro navigation, and a fine-grained map for navigation to the boundary of macro directions. So you have 2 small paths to calculate, and hence your search space is much much smaller than simply doing a single path to the destination. And if you're in the business of doing this a lot, you'd have a lot of that data pre-computed so at least part of the search is a search for pre-computed data, rather than a search for a path.

事实上，我已经做过很多次了，尝试了几种不同的方法。根据地图的大小(地理位置)，您可能会考虑使用haversine函数作为启发式方法。

我的最佳解决方案是使用带有直线距离的A*作为启发式函数。但接下来你需要地图上每个点(交集或顶点)的某种坐标。您还可以为启发式函数尝试不同的权重，即。

f(n) = k*h(n) + g(n)

k是一个大于0的常数。

我有点惊讶这里没有提到Floyd Warshall的算法。这个算法很像Dijkstra算法。它还有一个很好的特性，那就是它允许你计算，只要你想继续允许更多的中间顶点。因此，它自然会很快找到使用州际公路或高速公路的路线。

以下是世界上最快的路由算法的比较和正确性:

http://algo2.iti.uka.de/schultes/hwy/schultes_diss.pdf

下面是谷歌关于这个主题的技术演讲:

http://www.youtube.com/watch?v=-0ErpE8tQbw

以下是schultes所讨论的高速公路层次算法的实现(目前仅在柏林，我正在编写界面，移动版本也正在开发中):

http://tom.mapsforge.org/

我知道OP里的地图是怎么回事了:

用指定的中间点来观察路线:由于那条路不直，这条路线略微向后走。

如果他们的算法不会回溯，它就看不到更短的路线。

I was very curious about the heuristics used, when a while back we got routes from the same starting location near Santa Rosa, to two different campgrounds in Yosemite National Park. These different destinations produced quite different routes (via I-580 or CA-12) despite the fact that both routes converged for the last 100 miles (along CA-120) before diverging again by a few miles at the end. This was quite repeatable. The two routes were up to 50 miles apart for around 100 miles, but the distances/times were pretty close to each other as you would expect.

唉，我无法重现——算法肯定已经改变了。但这让我对算法很好奇。我所能推测的是，有一些方向修剪，恰好对从远处看的目的地之间的微小角度差异非常敏感，或者有不同的最终目的地选择的预先计算的片段。