如何计算C，C＃/ .NET 2.0或Java中所有情况下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

可能存在重复：

点和线段之间的最短距离

我正在寻找一种方法来计算所有情况下的最小距离。我发现的解决方案存在的问题是：

1. 图解概念图的解决方案始终显示垂直于线段的点，因此它介于线段的终点。我的几何技巧非常糟糕，所以我无法验证这些解决方案是否适用于所有情况。

算法解决方案是：使用fortran或其他语言不完全理解，
b：被人标记为不完整，c：调用没有以任何方式描述的方法/函数（被认为是微不足道的）。

2 a，b和c的好例子是

点与线段之间的最短距离

我有二维线段作为双类型坐标对（x1，y1），（x2，y2），并指向双类型坐标（x3，y3）。 C＃/ Java / C解决方案都值得赞赏。

感谢您的回答& BR：Matti

解决方案

也回答点与线段之间的最短距离，因为它收集所有语言的解决方案。答案也放在这里，因为这个问题具体要求C＃解决方案。这是从 http://www.topcoder.com/ tc？d1 = tutorials& d2 = geometry1& module = Static ：

  //计算点积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; 
} 
 
 //计算交叉产品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]; 
 
返回交叉; 
} 
 
 //计算从A到B 
的距离double距离（double [] pointA，double [] pointB）
 {
 double d1 = pointA [0]  -  pointB [0]; 
 double d2 = pointA [1]  -  pointB [1]; 
 
返回Math.Sqrt（d1 * d1 + d2 * d2）; 
} 
 
 //计算从AB到C 
的距离//如果isSegment为真，则AB是段而不是线。 
 double LineToPointDistance2D（double [] pointA，double [] pointB，double [] pointC，
 bool isSegment）
 {
 double dist = CrossProduct（pointA，pointB，pointC）/距离（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）; 
} 
返回Math.Abs​​（dist）; 
} 
  

Possible Duplicate:
Shortest distance between a point and a line segment

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

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.

2. 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).

Good example of 2 a, b and c is

Shortest distance between a point and a line segment

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.

Thanks for your answers & BR: Matti

解决方案

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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