我需要一个基本函数来求点到线段的最短距离。你可以随意用任何你想要的语言来编写解决方案;我可以把它翻译成什么我正在使用(Javascript)。
编辑:线段由两个端点定义。线段AB由两点A (x1,y1)和B (x2,y2)定义。我要求的是这条线段到点C (x3,y3)的距离。我的几何技能生疏了,所以我看到的例子让我很困惑,我很遗憾地承认。
我需要一个基本函数来求点到线段的最短距离。你可以随意用任何你想要的语言来编写解决方案;我可以把它翻译成什么我正在使用(Javascript)。
编辑:线段由两个端点定义。线段AB由两点A (x1,y1)和B (x2,y2)定义。我要求的是这条线段到点C (x3,y3)的距离。我的几何技能生疏了,所以我看到的例子让我很困惑,我很遗憾地承认。
当前回答
我制作了一个交互式Desmos图来演示如何实现这一点:
https://www.desmos.com/calculator/kswrm8ddum
红点是A点,绿点是B点,C点是蓝色点。 您可以拖动图形中的点来查看值的变化。 左边的值“s”是线段的参数(即s = 0表示点A, s = 1表示点B)。 值“d”是第三点到经过A和B的直线的距离。
编辑:
有趣的小见解:坐标(s, d)是坐标系中第三点C的坐标,AB是单位x轴,单位y轴垂直于AB。
其他回答
I'm assuming you want to find the shortest distance between the point and a line segment; to do this, you need to find the line (lineA) which is perpendicular to your line segment (lineB) which goes through your point, determine the intersection between that line (lineA) and your line which goes through your line segment (lineB); if that point is between the two points of your line segment, then the distance is the distance between your point and the point you just found which is the intersection of lineA and lineB; if the point is not between the two points of your line segment, you need to get the distance between your point and the closer of two ends of the line segment; this can be done easily by taking the square distance (to avoid a square root) between the point and the two points of the line segment; whichever is closer, take the square root of that one.
如果它是一条无限大的直线,而不是一条线段,最简单的方法是这样(在ruby中),其中mx + b是直线,(x1, y1)是已知的点
(y1 - mx1 - b).abs / Math.sqrt(m**2 + 1)
JavaScript中一个基于这个公式的更简洁的解决方案:
distToSegment: function (point, linePointA, linePointB){
var x0 = point.X;
var y0 = point.Y;
var x1 = linePointA.X;
var y1 = linePointA.Y;
var x2 = linePointB.X;
var y2 = linePointB.Y;
var Dx = (x2 - x1);
var Dy = (y2 - y1);
var numerator = Math.abs(Dy*x0 - Dx*y0 - x1*y2 + x2*y1);
var denominator = Math.sqrt(Dx*Dx + Dy*Dy);
if (denominator == 0) {
return this.dist2(point, linePointA);
}
return numerator/denominator;
}
这里没有看到Java实现,所以我将Javascript函数从接受的答案转换为Java代码:
static double sqr(double x) {
return x * x;
}
static double dist2(DoublePoint v, DoublePoint w) {
return sqr(v.x - w.x) + sqr(v.y - w.y);
}
static double distToSegmentSquared(DoublePoint p, DoublePoint v, DoublePoint w) {
double l2 = dist2(v, w);
if (l2 == 0) return dist2(p, v);
double t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / l2;
if (t < 0) return dist2(p, v);
if (t > 1) return dist2(p, w);
return dist2(p, new DoublePoint(
v.x + t * (w.x - v.x),
v.y + t * (w.y - v.y)
));
}
static double distToSegment(DoublePoint p, DoublePoint v, DoublePoint w) {
return Math.sqrt(distToSegmentSquared(p, v, w));
}
static class DoublePoint {
public double x;
public double y;
public DoublePoint(double x, double y) {
this.x = x;
this.y = y;
}
}
这里它使用Swift
/* Distance from a point (p1) to line l1 l2 */
func distanceFromPoint(p: CGPoint, toLineSegment l1: CGPoint, and l2: CGPoint) -> CGFloat {
let A = p.x - l1.x
let B = p.y - l1.y
let C = l2.x - l1.x
let D = l2.y - l1.y
let dot = A * C + B * D
let len_sq = C * C + D * D
let param = dot / len_sq
var xx, yy: CGFloat
if param < 0 || (l1.x == l2.x && l1.y == l2.y) {
xx = l1.x
yy = l1.y
} else if param > 1 {
xx = l2.x
yy = l2.y
} else {
xx = l1.x + param * C
yy = l1.y + param * D
}
let dx = p.x - xx
let dy = p.y - yy
return sqrt(dx * dx + dy * dy)
}