如何在 C、C#/.NET 2.0 或 Java 的所有情况下计算点和线段之间的最短 2D 距离? [英] how to calculate shortest 2D distance between a point and a line segment in all cases in C, C# / .NET 2.0 or Java?

问题描述

可能重复:
点到线段的最短距离

我正在寻找一种方法来计算所有情况下的最小距离.我发现的解决方案的问题是:

i am looking for a way to calculate the minimum distance in all cases. the problems with solutions i found are:

1. 具有图形概念图的解决方案显示点始终与线段垂直，因此它位于线段端点之间".我的几何技能很糟糕，所以我无法验证这些解决方案是否适用于所有情况.

1. Solutions with graphical conceptual drawings show point always on perpendicular from line segment so it's "between line segment's end points". My geometry skills are horrible so i can't verify that these solutions work in all cases.

算法解决方案是:使用 fortran 或我不完全理解的其他语言，b:被人们标记为不完整，c:调用未以任何方式描述的方法/函数(被认为是微不足道的).

Algorithm solutions are a: with fortran or some other language i don't fully understand, b: are flagged as incomplete by people, c: calling methods/functions that are not described in any way (considered trivial).

2 a、b、c 的好例子是

Good example of 2 a, b and c is

点到线段的最短距离

我将二维线段作为双坐标对 (x1, y1), (x2,y2) 并将点作为双坐标 (x3,y3).C#/Java/C 解决方案都受到赞赏.

i have the 2D line segment as double-type co-ordinate pair (x1, y1), (x2,y2) and point as double type co-ordinate (x3,y3). C#/Java/C solutions are all appreciated.

感谢您的回答&BR:马蒂

Thanks for your answers & BR: Matti

推荐答案

也回答了 点和线段之间的最短距离，因为它收集了所有语言的解决方案.答案也放在这里，因为这个问题专门询问 C# 解决方案.这是从 http://www.topcoder.com/修改的tc?d1=tutorials&d2=geometry1&module=Static :

Answered also Shortest distance between a point and a line segment because that gathers solutions in all languages. Answer put also here because this questions asks specifically a C# solution. This is modified from http://www.topcoder.com/tc?d1=tutorials&d2=geometry1&module=Static :

//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);
} 

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