查找折线/路径中的最近点

Question

我需要在GMSPolyline 的数组上找到距给定CLLocationCoordinate2D 最近的点。如果这样更好，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）用于此类计算？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

现在的问题是如何达到第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;
}

不过，问题出在精确点上。有什么算法吗？找到后我会在这里发布答案。

Answer 1

好的，我已经写好了。方法nearestPointToPoint:onLineSegmentPointA:pointB:distance: 允许您找到所选点和线段之间最近的坐标和距离（所以线与起点和终点）。

- (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（您只需要提供折线数组== self.polylines）并找到最好的。

// 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;
}

您可以使用它，例如：

- (void)mapView:(GMSMapView *)mapView didEndDraggingMarker:(GMSMarker *)marker {
    marker.position = [self nearestPolylineLocationToCoordinate:marker.position];
}

Answer 2

@Vive 的 nearestPointToPoint() 到 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)

但是在纬度 180 处出现了一些错误。我无法修复它。

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