newLatLngBounds(bounds, 50)的最后一行，50代表地图边缘和每个侧面上最外层标记之间的填充。public void centerIncidentRouteOnMap(List<LatLng> copiedPoints) {
double minLat = Integer.MAX_VALUE;
double maxLat = Integer.MIN_VALUE;
double minLon = Integer.MAX_VALUE;
double maxLon = Integer.MIN_VALUE;
for (LatLng point : copiedPoints) {
maxLat = Math.max(point.latitude, maxLat);
minLat = Math.min(point.latitude, minLat);
maxLon = Math.max(point.longitude, maxLon);
minLon = Math.min(point.longitude, minLon);
}
final LatLngBounds bounds = new LatLngBounds.Builder().include(new LatLng(maxLat, maxLon)).include(new LatLng(minLat, minLon)).build();
mapFragment.animateCamera(CameraUpdateFactory.newLatLngBounds(bounds, 50));
}
我发现之前的代码中有错误，于是决定重新编写。
我以前做过类似的事情，当时有大约4500个标记，并希望选择那些距离特定位置一定距离的标记。我将那段代码进行了泛化，使其适用于任何类型的标记。
我将要发布的代码包含两种方法：
selectLowDistanceMarkers
测量每个标记之间的距离，并仅选择那些与其他标记之间的距离不远的标记。由于每个标记之间的比较和之后的检查，这需要O(n+n^2)的运行时间。
getSurroundingMarkers
如果您已经知道要缩放到的位置，则此方法与上述方法相同。该方法的CPU负荷要小得多，因为它只需要对所有标记进行O(n)次运行并将它们与给定位置进行比较。
private List<Marker> selectLowDistanceMarkers(List<Marker> markers,
int maxDistanceMeters) {
List<Marker> acceptedMarkers = new ArrayList<Marker>();
if (markers == null) return acceptedMarkers;
Map<Marker, Float> longestDist = new HashMap<Marker, Float>();
for (Marker marker1 : markers) {
// in this for loop we remember the max distance for each marker
// think of a map with a flight company's routes from an airport
// these lines is drawn for each airport
// marker1 being the airport and marker2 destinations
for (Marker marker2 : markers) {
if (!marker1.equals(marker2)) {
float distance = distBetween(marker1.getPosition(),
marker2.getPosition());
if (longestDist.containsKey(marker1)) {
// possible we have a longer distance
if (distance > longestDist.get(marker1))
longestDist.put(marker1, distance);
} else {
// first distance
longestDist.put(marker1, distance);
}
}
}
}
// examine the distances collected
for (Marker marker: longestDist.keySet()) {
if (longestDist.get(marker) <= maxDistanceMeters) acceptedMarkers.add(marker);
}
return acceptedMarkers;
}
private List<Marker> getSurroundingMarkers(List<Marker> markers,
LatLng origin, int maxDistanceMeters) {
List<Marker> surroundingMarkers = surroundingMarkers = new ArrayList<Marker>();
if (markers == null) return surroundingMarkers ;
for (Marker marker : markers) {
double dist = distBetween(origin, marker.getPosition());
if (dist < getHydrantsLoadradius()) {
surroundingMarkers.add(marker);
}
}
return surroundingMarkers;
}
private float distBetween(LatLng pos1, LatLng pos2) {
return distBetween(pos1.latitude, pos1.longitude, pos2.latitude,
pos2.longitude);
}
/** distance in meters **/
private float distBetween(double lat1, double lng1, double lat2, double lng2) {
double earthRadius = 3958.75;
double dLat = Math.toRadians(lat2 - lat1);
double dLng = Math.toRadians(lng2 - lng1);
double a = Math.sin(dLat / 2) * Math.sin(dLat / 2)
+ Math.cos(Math.toRadians(lat1))
* Math.cos(Math.toRadians(lat2)) * Math.sin(dLng / 2)
* Math.sin(dLng / 2);
double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
double dist = earthRadius * c;
int meterConversion = 1609;
return (float) (dist * meterConversion);
}
在使用上述任何一种过滤算法后，再次使用众所周知的LatLngBounds来确定需要缩放的程度。
Display display = getWindowManager().getDefaultDisplay();
Point size = new Point();
display.getSize(size);
int ancho = size.x;
int alto =size.y;
List<LatLng> copiedPoints = new ArrayList<LatLng>();
copiedPoints.add(origin);
copiedPoints.add(dest);
centerIncidentRouteOnMap(copiedPoints, ancho, alto);
....
public void centerIncidentRouteOnMap(List<LatLng> copiedPoints, int ancho, int alto) {
double minLat = Integer.MAX_VALUE;
double maxLat = Integer.MIN_VALUE;
double minLon = Integer.MAX_VALUE;
double maxLon = Integer.MIN_VALUE;
for (LatLng point : copiedPoints) {
maxLat = Math.max(point.latitude, maxLat);
minLat = Math.min(point.latitude, minLat);
maxLon = Math.max(point.longitude, maxLon);
minLon = Math.min(point.longitude, minLon);
}
final LatLngBounds bounds = new LatLngBounds.Builder().include(new LatLng(maxLat, maxLon)).include(new LatLng(minLat, minLon)).build();
map.animateCamera(CameraUpdateFactory.newLatLngBounds(bounds,ancho, alto, 50));
}
基于cYrixmorten的一些想法，我简化了问题，因为我知道地图可以容纳至少4000公里的表面。所以这里是构建忽略网络摄像头列表的函数（然后我只需忽略该网络摄像头用于相机边界计算，但仍添加标记，以便在用户移动时出现在地图上）。
private List<Webcam> buildIgnoredWebcamsList(List<Webcam> webcams) {
if (webcams == null || webcams.size() < 2) return Lists.newArrayList();
int webcamCount = webcams.size();
// Number of conflicts (distance > 4000 km) for the camera at index #
float averageConflictCount = 0;
int[] conflictCount = new int[webcamCount];
Arrays.fill(conflictCount, 0);
// Find number of conflicts between camera pairs
float[] distance = new float[1];
for (int i = 0; i < webcamCount - 1; ++i) {
Webcam a = webcams.get(i);
// We don't have to start from 0, compare a and b only once
for (int j = i + 1; j < webcamCount; ++j) {
Webcam b = webcams.get(j);
Location.distanceBetween(a.getLatitude(), a.getLongitude(), b.getLatitude(), b.getLongitude(), distance);
// We have a conflict between a and b if they are more than 4000km away
if (distance[0] > 4000 * 1000) {
conflictCount[i] += 1;
conflictCount[j] += 1;
averageConflictCount += 2;
}
}
}
averageConflictCount /= webcamCount;
// Exclude all webcams with a number of conflicts greater than the average
List<Webcam> ignoredCamerasForBounds = Lists.newArrayList();
for (int i = 0; i < webcamCount; ++i) {
if (conflictCount[i] > averageConflictCount) {
ignoredCamerasForBounds.add(webcams.get(i));
}
}
return ignoredCamerasForBounds;
}