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
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)
其他回答
只是遇到了这个，我想我应该添加一个Lua实现。它假设点以表{x=xVal, y=yVal}给出，直线或线段由包含两个点的表给出(见下面的例子):
function distance( P1, P2 )
return math.sqrt((P1.x-P2.x)^2 + (P1.y-P2.y)^2)
end
-- Returns false if the point lies beyond the reaches of the segment
function distPointToSegment( line, P )
if line[1].x == line[2].x and line[1].y == line[2].y then
print("Error: Not a line!")
return false
end
local d = distance( line[1], line[2] )
local t = ((P.x - line[1].x)*(line[2].x - line[1].x) + (P.y - line[1].y)*(line[2].y - line[1].y))/(d^2)
local projection = {}
projection.x = line[1].x + t*(line[2].x-line[1].x)
projection.y = line[1].y + t*(line[2].y-line[1].y)
if t >= 0 and t <= 1 then -- within line segment?
return distance( projection, {x=P.x, y=P.y} )
else
return false
end
end
-- Returns value even if point is further down the line (outside segment)
function distPointToLine( line, P )
if line[1].x == line[2].x and line[1].y == line[2].y then
print("Error: Not a line!")
return false
end
local d = distance( line[1], line[2] )
local t = ((P.x - line[1].x)*(line[2].x - line[1].x) + (P.y - line[1].y)*(line[2].y - line[1].y))/(d^2)
local projection = {}
projection.x = line[1].x + t*(line[2].x-line[1].x)
projection.y = line[1].y + t*(line[2].y-line[1].y)
return distance( projection, {x=P.x, y=P.y} )
end
使用示例:
local P1 = {x = 0, y = 0}
local P2 = {x = 10, y = 10}
local line = { P1, P2 }
local P3 = {x = 7, y = 15}
print(distPointToLine( line, P3 )) -- prints 5.6568542494924
print(distPointToSegment( line, P3 )) -- prints false
这是我最后写的代码。这段代码假设一个点以{x:5, y:7}的形式定义。注意，这不是绝对最有效的方法，但它是我能想到的最简单、最容易理解的代码。
// a, b, and c in the code below are all points
function distance(a, b)
{
var dx = a.x - b.x;
var dy = a.y - b.y;
return Math.sqrt(dx*dx + dy*dy);
}
function Segment(a, b)
{
var ab = {
x: b.x - a.x,
y: b.y - a.y
};
var length = distance(a, b);
function cross(c) {
return ab.x * (c.y-a.y) - ab.y * (c.x-a.x);
};
this.distanceFrom = function(c) {
return Math.min(distance(a,c),
distance(b,c),
Math.abs(cross(c) / length));
};
}
这个答案是基于公认答案的JavaScript解决方案。 它主要只是格式更好，函数名更长，当然函数语法更短，因为它是在ES6 + CoffeeScript中。
JavaScript版本(ES6)
distanceSquared = (v, w)=> Math.pow(v.x - w.x, 2) + Math.pow(v.y - w.y, 2);
distance = (v, w)=> Math.sqrt(distanceSquared(v, w));
distanceToLineSegmentSquared = (p, v, w)=> {
l2 = distanceSquared(v, w);
if (l2 === 0) {
return distanceSquared(p, v);
}
t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / l2;
t = Math.max(0, Math.min(1, t));
return distanceSquared(p, {
x: v.x + t * (w.x - v.x),
y: v.y + t * (w.y - v.y)
});
}
distanceToLineSegment = (p, v, w)=> {
return Math.sqrt(distanceToLineSegmentSquared(p, v));
}
CoffeeScript版本
distanceSquared = (v, w)-> (v.x - w.x) ** 2 + (v.y - w.y) ** 2
distance = (v, w)-> Math.sqrt(distanceSquared(v, w))
distanceToLineSegmentSquared = (p, v, w)->
l2 = distanceSquared(v, w)
return distanceSquared(p, v) if l2 is 0
t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / l2
t = Math.max(0, Math.min(1, t))
distanceSquared(p, {
x: v.x + t * (w.x - v.x)
y: v.y + t * (w.y - v.y)
})
distanceToLineSegment = (p, v, w)->
Math.sqrt(distanceToLineSegmentSquared(p, v, w))
WPF版本:
public class LineSegment
{
private readonly Vector _offset;
private readonly Vector _vector;
public LineSegment(Point start, Point end)
{
_offset = (Vector)start;
_vector = (Vector)(end - _offset);
}
public double DistanceTo(Point pt)
{
var v = (Vector)pt - _offset;
// first, find a projection point on the segment in parametric form (0..1)
var p = (v * _vector) / _vector.LengthSquared;
// and limit it so it lays inside the segment
p = Math.Min(Math.Max(p, 0), 1);
// now, find the distance from that point to our point
return (_vector * p - v).Length;
}
}
在我自己的问题线程如何计算在C, c# / .NET 2.0或Java的所有情况下一个点和线段之间的最短2D距离?当我找到一个c#的答案时，我被要求把它放在这里:所以它是从http://www.topcoder.com/tc?d1=tutorials&d2=geometry1&module=Static修改的:
//Compute the dot product AB . BC
private double DotProduct(double[] pointA, double[] pointB, double[] pointC)
{
double[] AB = new double[2];
double[] BC = new double[2];
AB[0] = pointB[0] - pointA[0];
AB[1] = pointB[1] - pointA[1];
BC[0] = pointC[0] - pointB[0];
BC[1] = pointC[1] - pointB[1];
double dot = AB[0] * BC[0] + AB[1] * BC[1];
return dot;
}
//Compute the cross product AB x AC
private double CrossProduct(double[] pointA, double[] pointB, double[] pointC)
{
double[] AB = new double[2];
double[] AC = new double[2];
AB[0] = pointB[0] - pointA[0];
AB[1] = pointB[1] - pointA[1];
AC[0] = pointC[0] - pointA[0];
AC[1] = pointC[1] - pointA[1];
double cross = AB[0] * AC[1] - AB[1] * AC[0];
return cross;
}
//Compute the distance from A to B
double Distance(double[] pointA, double[] pointB)
{
double d1 = pointA[0] - pointB[0];
double d2 = pointA[1] - pointB[1];
return Math.Sqrt(d1 * d1 + d2 * d2);
}
//Compute the distance from AB to C
//if isSegment is true, AB is a segment, not a line.
double LineToPointDistance2D(double[] pointA, double[] pointB, double[] pointC,
bool isSegment)
{
double dist = CrossProduct(pointA, pointB, pointC) / Distance(pointA, pointB);
if (isSegment)
{
double dot1 = DotProduct(pointA, pointB, pointC);
if (dot1 > 0)
return Distance(pointB, pointC);
double dot2 = DotProduct(pointB, pointA, pointC);
if (dot2 > 0)
return Distance(pointA, pointC);
}
return Math.Abs(dist);
}
我不是要回答问题，而是要问问题，所以我希望我不会因为某些原因而得到数百万张反对票，而是批评。我只是想(并被鼓励)分享其他人的想法，因为这个帖子中的解决方案要么是用一些奇异的语言(Fortran, Mathematica)，要么被某人标记为错误。对我来说唯一有用的(由Grumdrig编写)是用c++编写的，没有人标记它有错误。但是它缺少被调用的方法(dot等)。
