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
C#
改编自@Grumdrig
public static double MinimumDistanceToLineSegment(this Point p,
Line line)
{
var v = line.StartPoint;
var w = line.EndPoint;
double lengthSquared = DistanceSquared(v, w);
if (lengthSquared == 0.0)
return Distance(p, v);
double t = Math.Max(0, Math.Min(1, DotProduct(p - v, w - v) / lengthSquared));
var projection = v + t * (w - v);
return Distance(p, projection);
}
public static double Distance(Point a, Point b)
{
return Math.Sqrt(DistanceSquared(a, b));
}
public static double DistanceSquared(Point a, Point b)
{
var d = a - b;
return DotProduct(d, d);
}
public static double DotProduct(Point a, Point b)
{
return (a.X * b.X) + (a.Y * b.Y);
}
其他回答
这是Javascript中最简单的完整代码。
(X, y)是目标点(x1, y)到(x2, y)是线段。
更新:修复了评论中0长度的行问题。
function pDistance(x, y, x1, y1, x2, y2) {
var A = x - x1;
var B = y - y1;
var C = x2 - x1;
var D = y2 - y1;
var dot = A * C + B * D;
var len_sq = C * C + D * D;
var param = -1;
if (len_sq != 0) //in case of 0 length line
param = dot / len_sq;
var xx, yy;
if (param < 0) {
xx = x1;
yy = y1;
}
else if (param > 1) {
xx = x2;
yy = y2;
}
else {
xx = x1 + param * C;
yy = y1 + param * D;
}
var dx = x - xx;
var dy = y - yy;
return Math.sqrt(dx * dx + dy * dy);
}
更新:Kotlin版本
fun getDistance(x: Double, y: Double, x1: Double, y1: Double, x2: Double, y2: Double): Double {
val a = x - x1
val b = y - y1
val c = x2 - x1
val d = y2 - y1
val lenSq = c * c + d * d
val param = if (lenSq != .0) { //in case of 0 length line
val dot = a * c + b * d
dot / lenSq
} else {
-1.0
}
val (xx, yy) = when {
param < 0 -> x1 to y1
param > 1 -> x2 to y2
else -> x1 + param * c to y1 + param * d
}
val dx = x - xx
val dy = y - yy
return hypot(dx, dy)
}
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)
}
上面的函数在垂直线上不起作用。这是一个工作正常的函数! 与点p1 p2相交。CheckPoint为p;
public float DistanceOfPointToLine2(PointF p1, PointF p2, PointF p)
{
// (y1-y2)x + (x2-x1)y + (x1y2-x2y1)
//d(P,L) = --------------------------------
// sqrt( (x2-x1)pow2 + (y2-y1)pow2 )
double ch = (p1.Y - p2.Y) * p.X + (p2.X - p1.X) * p.Y + (p1.X * p2.Y - p2.X * p1.Y);
double del = Math.Sqrt(Math.Pow(p2.X - p1.X, 2) + Math.Pow(p2.Y - p1.Y, 2));
double d = ch / del;
return (float)d;
}
一个2D和3D的解决方案
考虑基底的变化,使得线段变成(0,0,0)-(d, 0,0)和点(u, v, 0)。在这个平面上,最短的距离由
u ≤ 0 -> d(A, C)
0 ≤ u ≤ d -> |v|
d ≤ u -> d(B, C)
(到其中一个端点或到支撑线的距离,取决于到该线的投影。等距轨迹由两个半圆和两条线段组成。)
式中,d为AB线段的长度,u、v分别为AB/d (AB方向的单位矢量)与AC的标量积和外积的模量。
AB.AC ≤ 0 -> |AC|
0 ≤ AB.AC ≤ AB² -> |ABxAC|/|AB|
AB² ≤ AB.AC -> |BC|
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;
}
}