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
请参见以下网站中的Matlab几何工具箱: http://people.sc.fsu.edu/~jburkardt/m_src/geometry/geometry.html
按Ctrl +f,输入“segment”,查找线段相关函数。函数“segment_point_dist_2d.”和segment_point_dist_3d。M "是你需要的。
几何代码有C版本、c++版本、FORTRAN77版本、FORTRAN90版本和MATLAB版本。
其他回答
GLSL版:
// line (a -> b ) point p[enter image description here][1]
float distanceToLine(vec2 a, vec2 b, vec2 p) {
float aside = dot((p - a),(b - a));
if(aside< 0.0) return length(p-a);
float bside = dot((p - b),(a - b));
if(bside< 0.0) return length(p-b);
vec2 pointOnLine = (bside*a + aside*b)/pow(length(a-b),2.0);
return length(p - pointOnLine);
}
这里是与c++答案相同的东西,但移植到pascal。点参数的顺序已经改变,以适应我的代码,但还是一样的东西。
function Dot(const p1, p2: PointF): double;
begin
Result := p1.x * p2.x + p1.y * p2.y;
end;
function SubPoint(const p1, p2: PointF): PointF;
begin
result.x := p1.x - p2.x;
result.y := p1.y - p2.y;
end;
function ShortestDistance2(const p,v,w : PointF) : double;
var
l2,t : double;
projection,tt: PointF;
begin
// Return minimum distance between line segment vw and point p
//l2 := length_squared(v, w); // i.e. |w-v|^2 - avoid a sqrt
l2 := Distance(v,w);
l2 := MPower(l2,2);
if (l2 = 0.0) then begin
result:= Distance(p, v); // v == w case
exit;
end;
// Consider the line extending the segment, parameterized as v + t (w - v).
// We find projection of point p onto the line.
// It falls where t = [(p-v) . (w-v)] / |w-v|^2
t := Dot(SubPoint(p,v),SubPoint(w,v)) / l2;
if (t < 0.0) then begin
result := Distance(p, v); // Beyond the 'v' end of the segment
exit;
end
else if (t > 1.0) then begin
result := Distance(p, w); // Beyond the 'w' end of the segment
exit;
end;
//projection := v + t * (w - v); // Projection falls on the segment
tt.x := v.x + t * (w.x - v.x);
tt.y := v.y + t * (w.y - v.y);
result := Distance(p, tt);
end;
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;
}
}
和这个答案一样,只是用的是Visual Basic。使其可作为Microsoft Excel和VBA/宏中的用户定义函数使用。
函数返回点(x,y)到由(x1,y1)和(x2,y2)定义的线段的最近距离。
Function DistanceToSegment(x As Double, y As Double, x1 As Double, y1 As Double, x2 As Double, y2 As Double)
Dim A As Double
A = x - x1
Dim B As Double
B = y - y1
Dim C As Double
C = x2 - x1
Dim D As Double
D = y2 - y1
Dim dot As Double
dot = A * C + B * D
Dim len_sq As Double
len_sq = C * C + D * D
Dim param As Double
param = -1
If (len_sq <> 0) Then
param = dot / len_sq
End If
Dim xx As Double
Dim yy As Double
If (param < 0) Then
xx = x1
yy = y1
ElseIf (param > 1) Then
xx = x2
yy = y2
Else
xx = x1 + param * C
yy = y1 + param * D
End If
Dim dx As Double
dx = x - xx
Dim dy As Double
dy = y - yy
DistanceToSegment = Math.Sqr(dx * dx + dy * dy)
End Function
对于懒人来说,以下是我在Objective-C语言中移植@Grumdrig的解决方案:
CGFloat sqr(CGFloat x) { return x*x; }
CGFloat dist2(CGPoint v, CGPoint w) { return sqr(v.x - w.x) + sqr(v.y - w.y); }
CGFloat distanceToSegmentSquared(CGPoint p, CGPoint v, CGPoint w)
{
CGFloat l2 = dist2(v, w);
if (l2 == 0.0f) return dist2(p, v);
CGFloat t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / l2;
if (t < 0.0f) return dist2(p, v);
if (t > 1.0f) return dist2(p, w);
return dist2(p, CGPointMake(v.x + t * (w.x - v.x), v.y + t * (w.y - v.y)));
}
CGFloat distanceToSegment(CGPoint point, CGPoint segmentPointV, CGPoint segmentPointW)
{
return sqrtf(distanceToSegmentSquared(point, segmentPointV, segmentPointW));
}