我想知道Google地图用哪种算法计算两点之间的方向?Google是否曾经提到过它?顺便说一下:我正在询问Google用于查找两点之间最短路线的算法。
简而言之...
Hub标签算法为静态道路网络提供了最快的查询速度,但需要大量内存运行(18 GiB)。
过渡节点路由略慢,但只需要约2 GiB的内存,并具有更快的预处理时间。
收缩分层提供了快速预处理时间、低空间要求(0.4 GiB)和快速查询时间之间的良好平衡。
没有一种算法完全占优势...
这个由Peter Sanders主讲的Google技术讲座可能会引起您的兴趣。https://www.youtube.com/watch?v=-0ErpE8tQbw
还有这个由安德鲁·戈德伯格发表的演讲
https://www.youtube.com/watch?v=WPrkc78XLhw
可以从KIT的Peter Sanders研究小组网站上获取收缩分层的开源实现。http://algo2.iti.kit.edu/english/routeplanning.php
如果你指的是谷歌地图方向API和两点之间的最短路径,那么它是一个图论问题,可以使用Dijkstra算法解决。这是一个带有回溯的DFS。
像这样的疑问,您应该始终检查Android源代码。
private static void computeDistanceAndBearing(double lat1, double lon1,
double lat2, double lon2, float[] results) {
int MAXITERS = 20;
// Convert lat/long to radians
lat1 *= Math.PI / 180.0;
lat2 *= Math.PI / 180.0;
lon1 *= Math.PI / 180.0;
lon2 *= Math.PI / 180.0;
double a = 6378137.0; // WGS84 major axis
double b = 6356752.3142; // WGS84 semi-major axis
double f = (a - b) / a;
double aSqMinusBSqOverBSq = (a * a - b * b) / (b * b);
double L = lon2 - lon1;
double A = 0.0;
double U1 = Math.atan((1.0 - f) * Math.tan(lat1));
double U2 = Math.atan((1.0 - f) * Math.tan(lat2));
double cosU1 = Math.cos(U1);
double cosU2 = Math.cos(U2);
double sinU1 = Math.sin(U1);
double sinU2 = Math.sin(U2);
double cosU1cosU2 = cosU1 * cosU2;
double sinU1sinU2 = sinU1 * sinU2;
double sigma = 0.0;
double deltaSigma = 0.0;
double cosSqAlpha = 0.0;
double cos2SM = 0.0;
double cosSigma = 0.0;
double sinSigma = 0.0;
double cosLambda = 0.0;
double sinLambda = 0.0;
double lambda = L; // initial guess
for (int iter = 0; iter < MAXITERS; iter++) {
double lambdaOrig = lambda;
cosLambda = Math.cos(lambda);
sinLambda = Math.sin(lambda);
double t1 = cosU2 * sinLambda;
double t2 = cosU1 * sinU2 - sinU1 * cosU2 * cosLambda;
double sinSqSigma = t1 * t1 + t2 * t2; // (14)
sinSigma = Math.sqrt(sinSqSigma);
cosSigma = sinU1sinU2 + cosU1cosU2 * cosLambda; // (15)
sigma = Math.atan2(sinSigma, cosSigma); // (16)
double sinAlpha = (sinSigma == 0) ? 0.0 :
cosU1cosU2 * sinLambda / sinSigma; // (17)
cosSqAlpha = 1.0 - sinAlpha * sinAlpha;
cos2SM = (cosSqAlpha == 0) ? 0.0 :
cosSigma - 2.0 * sinU1sinU2 / cosSqAlpha; // (18)
double uSquared = cosSqAlpha * aSqMinusBSqOverBSq; // defn
A = 1 + (uSquared / 16384.0) * // (3)
(4096.0 + uSquared *
(-768 + uSquared * (320.0 - 175.0 * uSquared)));
double B = (uSquared / 1024.0) * // (4)
(256.0 + uSquared *
(-128.0 + uSquared * (74.0 - 47.0 * uSquared)));
double C = (f / 16.0) *
cosSqAlpha *
(4.0 + f * (4.0 - 3.0 * cosSqAlpha)); // (10)
double cos2SMSq = cos2SM * cos2SM;
deltaSigma = B * sinSigma * // (6)
(cos2SM + (B / 4.0) *
(cosSigma * (-1.0 + 2.0 * cos2SMSq) -
(B / 6.0) * cos2SM *
(-3.0 + 4.0 * sinSigma * sinSigma) *
(-3.0 + 4.0 * cos2SMSq)));
lambda = L +
(1.0 - C) * f * sinAlpha *
(sigma + C * sinSigma *
(cos2SM + C * cosSigma *
(-1.0 + 2.0 * cos2SM * cos2SM))); // (11)
double delta = (lambda - lambdaOrig) / lambda;
if (Math.abs(delta) < 1.0e-12) {
break;
}
}
float distance = (float) (b * A * (sigma - deltaSigma));
results[0] = distance;
if (results.length > 1) {
float initialBearing = (float) Math.atan2(cosU2 * sinLambda,
cosU1 * sinU2 - sinU1 * cosU2 * cosLambda);
initialBearing *= 180.0 / Math.PI;
results[1] = initialBearing;
if (results.length > 2) {
float finalBearing = (float) Math.atan2(cosU1 * sinLambda,
-sinU1 * cosU2 + cosU1 * sinU2 * cosLambda);
finalBearing *= 180.0 / Math.PI;
results[2] = finalBearing;
}
}
}
谷歌地图 API 中的几何库提供了算法,你可以在源代码中找到它。
我不确定谷歌地图是否使用相同的算法。
这个算法很简单:
function toRadians(deg){
return deg * (Math.PI / 180);
}
function getDistance(from, to) {
var c = toRadians(from.lat()),
d = toRadians(to.lat());
return 2 * Math.asin(Math.sqrt(Math.pow(Math.sin((c - d) / 2), 2) + Math.cos(c) * Math.cos(d) * Math.pow(Math.sin((toRadians(from.lng()) - toRadians(to.lng())) / 2), 2))) * 6378137;
}
这两行代码将会得到相同的结果:
console.log(google.maps.geometry.spherical.computeDistanceBetween(new google.maps.LatLng(39.915, 116.404), new google.maps.LatLng(38.8871, 113.3113)));
console.log(getDistance(new google.maps.LatLng(39.915, 116.404), new google.maps.LatLng(38.8871, 113.3113)));