我需要找到CLLocationCoordinate2D数组上给定GMSPolyline的最近点。如果那更好，我可以将其转换为GMSPath。是否有任何现成的方法（或任何存储库）进行此类计算？我在实施方面遇到了一些问题。我想知道如何创建算法：
1. for all polylines
1.1. find smallest distance between polyline and touch point, save CLLocationCoordinate2D
2. for all distances from point 1.1.
2.1. find the shortest one, it's CLLocationCoordinate2D is our point
现在的问题是如何实现第1.1点......？
基于SOF shortest distance question，我写了这样的代码：
- (void)findNearestLineSegmentToCoordinate:(CLLocationCoordinate2D)coordinate {
GMSPolyline *bestPolyline;
double bestDistance = DBL_MAX;
CGPoint originPoint = CGPointMake(coordinate.longitude, coordinate.latitude);
for (GMSPolyline *polyline in self.polylines) {
polyline.strokeColor = [UIColor redColor]; // TMP
if (polyline.path.count < 2) { // we need at least 2 points: start and end
return;
}
for (NSInteger index = 0; index < polyline.path.count - 1; index++) {
CLLocationCoordinate2D startCoordinate = [polyline.path coordinateAtIndex:index];
CGPoint startPoint = CGPointMake(startCoordinate.longitude, startCoordinate.latitude);
CLLocationCoordinate2D endCoordinate = [polyline.path coordinateAtIndex:(index + 1)];
CGPoint endPoint = CGPointMake(endCoordinate.longitude, endCoordinate.latitude);
double distance = [self distanceToPoint:originPoint fromLineSegmentBetween:startPoint and:endPoint];
if (distance < bestDistance) {
bestDistance = distance;
bestPolyline = polyline;
}
}
}
bestPolyline.map = nil;
bestPolyline.strokeColor = [UIColor greenColor]; // TMP
bestPolyline.map = self.aView.mapView;
}
但问题仍然存在。任何算法？我发现时会在这里发布答案。
答案 0 :(得分：6)
好的，我已经设法写了。方法nearestPointToPoint:onLineSegmentPointA:pointB:distance:允许您同时找到所选点和线段之间的最近坐标和距离（因此，以开头和结尾排列）。
- (CLLocationCoordinate2D)nearestPolylineLocationToCoordinate:(CLLocationCoordinate2D)coordinate {
GMSPolyline *bestPolyline;
double bestDistance = DBL_MAX;
CGPoint bestPoint;
CGPoint originPoint = CGPointMake(coordinate.longitude, coordinate.latitude);
for (GMSPolyline *polyline in self.polylines) {
if (polyline.path.count < 2) { // we need at least 2 points: start and end
return kCLLocationCoordinate2DInvalid;
}
for (NSInteger index = 0; index < polyline.path.count - 1; index++) {
CLLocationCoordinate2D startCoordinate = [polyline.path coordinateAtIndex:index];
CGPoint startPoint = CGPointMake(startCoordinate.longitude, startCoordinate.latitude);
CLLocationCoordinate2D endCoordinate = [polyline.path coordinateAtIndex:(index + 1)];
CGPoint endPoint = CGPointMake(endCoordinate.longitude, endCoordinate.latitude);
double distance;
CGPoint point = [self nearestPointToPoint:originPoint onLineSegmentPointA:startPoint pointB:endPoint distance:&distance];
if (distance < bestDistance) {
bestDistance = distance;
bestPolyline = polyline;
bestPoint = point;
}
}
}
return CLLocationCoordinate2DMake(bestPoint.y, bestPoint.x);
}
方法nearestPolylineLocationToCoordinate:将浏览所有折线（您只需提供折线数组== self.polylines）并找到最佳折线。
// taken and modified from: http://stackoverflow.com/questions/849211/shortest-distance-between-a-point-and-a-line-segment
- (CGPoint)nearestPointToPoint:(CGPoint)origin onLineSegmentPointA:(CGPoint)pointA pointB:(CGPoint)pointB distance:(double *)distance {
CGPoint dAP = CGPointMake(origin.x - pointA.x, origin.y - pointA.y);
CGPoint dAB = CGPointMake(pointB.x - pointA.x, pointB.y - pointA.y);
CGFloat dot = dAP.x * dAB.x + dAP.y * dAB.y;
CGFloat squareLength = dAB.x * dAB.x + dAB.y * dAB.y;
CGFloat param = dot / squareLength;
CGPoint nearestPoint;
if (param < 0 || (pointA.x == pointB.x && pointA.y == pointB.y)) {
nearestPoint.x = pointA.x;
nearestPoint.y = pointA.y;
} else if (param > 1) {
nearestPoint.x = pointB.x;
nearestPoint.y = pointB.y;
} else {
nearestPoint.x = pointA.x + param * dAB.x;
nearestPoint.y = pointA.y + param * dAB.y;
}
CGFloat dx = origin.x - nearestPoint.x;
CGFloat dy = origin.y - nearestPoint.y;
*distance = sqrtf(dx * dx + dy * dy);
return nearestPoint;
}
您可以在例如：
中使用它- (void)mapView:(GMSMapView *)mapView didEndDraggingMarker:(GMSMarker *)marker {
marker.position = [self nearestPolylineLocationToCoordinate:marker.position];
}
答案 1 :(得分：0)
@Vive的nearestPointToPoint()至Swift 5
func nearestPointToPoint(_ origin: CGPoint, _ pointA: CGPoint, _ pointB: CGPoint) -> (CGPoint, Double) {
let dAP = CGPoint(x: origin.x - pointA.x, y: origin.y - pointA.y)
let dAB = CGPoint(x: pointB.x - pointA.x, y: pointB.y - pointA.y)
let dot = dAP.x * dAB.x + dAP.y * dAB.y
let squareLength = dAB.x * dAB.x + dAB.y * dAB.y
let param = dot / squareLength
// abnormal value at near latitude 180
// var nearestPoint = CGPoint()
// if param < 0 || (pointA.x == pointB.x && pointA.y == pointB.y) {
// nearestPoint.x = pointA.x
// nearestPoint.y = pointA.y
// } else if param > 1 {
// nearestPoint.x = pointB.x
// nearestPoint.y = pointB.y
// } else {
// nearestPoint.x = pointA.x + param * dAB.x
// nearestPoint.y = pointA.y + param * dAB.y
// }
let nearestPoint = CGPoint(x: pointA.x + param * dAB.x,
y: pointA.y + param * dAB.y)
let dx = origin.x - nearestPoint.x
let dy = origin.y - nearestPoint.y
let distance = sqrtf(Float(dx * dx + dy * dy))
return (nearestPoint, Double(distance))
}
结果是
me: (53, -1), pointA: (52, -2), pointB(52, 0) => (52, -1)
me: (53, 0), pointA: (52, -1), pointB(52, 1) => (52, 0)
me: (53, 1), pointA: (52, 0), pointB(52, 2) => (52, 1)
me: (-1, -77), pointA: (0, -78), pointB(-2, -78) => (-1, -78)
me: (0, -77), pointA: (-1, -78), pointB(1, -78) => (0, -78)
me: (1, -77), pointA: (2, -78), pointB(0, -78) => (1, -78)
me: (1, 179), pointA: (0, 178), pointB(0, 180) => (0, 179)
me: (1, 180), pointA: (0, 179), pointB(0, -179) => (0, 180)
me: (1, -179), pointA: (0, 180), pointB(0, -178) => (0, -179)
但是在北纬180度出现一些错误。我无法解决。
me: (0, 180), pointA: (0, 179), pointB(1, 180) => (0.5, 179.5)
me: (0, -179), pointA: (0, 179), pointB(1, -179) => (0.99, -178.99) // abnormal