我正在开发一个能够追踪谷歌地图SDK Add-On上公路的应用程序。
有许多点被描绘,无意中使地图非常简练。 看起来,SOS地图会关注这一问题,因为如果不绘制地图,就可以绘制出大量要点。
我开始采样点,取决于动物群。 例如,7级动物群只使用并沿用每20个点,以追踪路线。 但是,即便如此,也仍然太多,其结果却充满了非常缓慢的pan积和om淡的经验。
两个问题:
一种:能否发现电话中显示的地图的目前区域,再不设掩饰点?
<>strong>EDIT——经过一些初步调查,很容易发现剪辑区,而不是油漆点/线。 quickReject是用于这一目的的I ve。 由于所有要点仍在探讨之中,以确定是否必须加以利用,这无助于业绩的取得——我相信,可以优化。
两点:由于对要点的预测,是否完全有可能打脚,或许是走了这条道路(必须做的是每个动物群级的工作)。
能否探测到电话中显示的地图的目前区域,再不要在屏幕上放置油漆点?
是的,你可以预测你的屏幕四角:
Projection proj = mapView.getProjection();
GeoPoint topLeft = proj.fromPixels(0, 0);
GeoPoint bottomRight = proj.fromPixels(mapView.getWidth()-1, mapView.getHeight()-1);
double topLat = topLeft.getLatitudeE6()/1E6;
double topLon = topLeft.getLongitudeE6()/1E6;
double bottomLat = bottomRight.getLatitudeE6()/1E6;
double bottomLon = bottomRight.getLongitudeE6()/1E6;
在作出这一规定时,你只应选择这些界限中的点。
另见:
如果你处于非常高的血清水平,就会发现:
个人使用:outeeuptetics,以减少直线的点数。 例如,如果20点位于同一线,就应当使用起始点和终点点,并划定一条直线。
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