我有两个点（一个线段）和一个矩形。我想知道如何计算线段是否与矩形相交。
答案 0 :(得分：21)
来自我的“几何”课程：
public struct Line
{
public static Line Empty;
private PointF p1;
private PointF p2;
public Line(PointF p1, PointF p2)
{
this.p1 = p1;
this.p2 = p2;
}
public PointF P1
{
get { return p1; }
set { p1 = value; }
}
public PointF P2
{
get { return p2; }
set { p2 = value; }
}
public float X1
{
get { return p1.X; }
set { p1.X = value; }
}
public float X2
{
get { return p2.X; }
set { p2.X = value; }
}
public float Y1
{
get { return p1.Y; }
set { p1.Y = value; }
}
public float Y2
{
get { return p2.Y; }
set { p2.Y = value; }
}
}
public struct Polygon: IEnumerable<PointF>
{
private PointF[] points;
public Polygon(PointF[] points)
{
this.points = points;
}
public PointF[] Points
{
get { return points; }
set { points = value; }
}
public int Length
{
get { return points.Length; }
}
public PointF this[int index]
{
get { return points[index]; }
set { points[index] = value; }
}
public static implicit operator PointF[](Polygon polygon)
{
return polygon.points;
}
public static implicit operator Polygon(PointF[] points)
{
return new Polygon(points);
}
IEnumerator<PointF> IEnumerable<PointF>.GetEnumerator()
{
return (IEnumerator<PointF>)points.GetEnumerator();
}
public IEnumerator GetEnumerator()
{
return points.GetEnumerator();
}
}
public enum Intersection
{
None,
Tangent,
Intersection,
Containment
}
public static class Geometry
{
public static Intersection IntersectionOf(Line line, Polygon polygon)
{
if (polygon.Length == 0)
{
return Intersection.None;
}
if (polygon.Length == 1)
{
return IntersectionOf(polygon[0], line);
}
bool tangent = false;
for (int index = 0; index < polygon.Length; index++)
{
int index2 = (index + 1)%polygon.Length;
Intersection intersection = IntersectionOf(line, new Line(polygon[index], polygon[index2]));
if (intersection == Intersection.Intersection)
{
return intersection;
}
if (intersection == Intersection.Tangent)
{
tangent = true;
}
}
return tangent ? Intersection.Tangent : IntersectionOf(line.P1, polygon);
}
public static Intersection IntersectionOf(PointF point, Polygon polygon)
{
switch (polygon.Length)
{
case 0:
return Intersection.None;
case 1:
if (polygon[0].X == point.X && polygon[0].Y == point.Y)
{
return Intersection.Tangent;
}
else
{
return Intersection.None;
}
case 2:
return IntersectionOf(point, new Line(polygon[0], polygon[1]));
}
int counter = 0;
int i;
PointF p1;
int n = polygon.Length;
p1 = polygon[0];
if (point == p1)
{
return Intersection.Tangent;
}
for (i = 1; i <= n; i++)
{
PointF p2 = polygon[i % n];
if (point == p2)
{
return Intersection.Tangent;
}
if (point.Y > Math.Min(p1.Y, p2.Y))
{
if (point.Y <= Math.Max(p1.Y, p2.Y))
{
if (point.X <= Math.Max(p1.X, p2.X))
{
if (p1.Y != p2.Y)
{
double xinters = (point.Y - p1.Y) * (p2.X - p1.X) / (p2.Y - p1.Y) + p1.X;
if (p1.X == p2.X || point.X <= xinters)
counter++;
}
}
}
}
p1 = p2;
}
return (counter % 2 == 1) ? Intersection.Containment : Intersection.None;
}
public static Intersection IntersectionOf(PointF point, Line line)
{
float bottomY = Math.Min(line.Y1, line.Y2);
float topY = Math.Max(line.Y1, line.Y2);
bool heightIsRight = point.Y >= bottomY &&
point.Y <= topY;
//Vertical line, slope is divideByZero error!
if (line.X1 == line.X2)
{
if (point.X == line.X1 && heightIsRight)
{
return Intersection.Tangent;
}
else
{
return Intersection.None;
}
}
float slope = (line.X2 - line.X1)/(line.Y2 - line.Y1);
bool onLine = (line.Y1 - point.Y) == (slope*(line.X1 - point.X));
if (onLine && heightIsRight)
{
return Intersection.Tangent;
}
else
{
return Intersection.None;
}
}
}
答案 1 :(得分：3)
对于线条和矩形的每一边执行http://mathworld.wolfram.com/Line-LineIntersection.html 或者：http://mathworld.wolfram.com/Line-PlaneIntersection.html
答案 2 :(得分：2)
因为它缺失了所以我只是为了完整性添加它
public static Intersection IntersectionOf(Line line1, Line line2)
{
// Fail if either line segment is zero-length.
if (line1.X1 == line1.X2 && line1.Y1 == line1.Y2 || line2.X1 == line2.X2 && line2.Y1 == line2.Y2)
return Intersection.None;
if (line1.X1 == line2.X1 && line1.Y1 == line2.Y1 || line1.X2 == line2.X1 && line1.Y2 == line2.Y1)
return Intersection.Intersection;
if (line1.X1 == line2.X2 && line1.Y1 == line2.Y2 || line1.X2 == line2.X2 && line1.Y2 == line2.Y2)
return Intersection.Intersection;
// (1) Translate the system so that point A is on the origin.
line1.X2 -= line1.X1; line1.Y2 -= line1.Y1;
line2.X1 -= line1.X1; line2.Y1 -= line1.Y1;
line2.X2 -= line1.X1; line2.Y2 -= line1.Y1;
// Discover the length of segment A-B.
double distAB = Math.Sqrt(line1.X2 * line1.X2 + line1.Y2 * line1.Y2);
// (2) Rotate the system so that point B is on the positive X axis.
double theCos = line1.X2 / distAB;
double theSin = line1.Y2 / distAB;
double newX = line2.X1 * theCos + line2.Y1 * theSin;
line2.Y1 = line2.Y1 * theCos - line2.X1 * theSin; line2.X1 = newX;
newX = line2.X2 * theCos + line2.Y2 * theSin;
line2.Y2 = line2.Y2 * theCos - line2.X2 * theSin; line2.X2 = newX;
// Fail if segment C-D doesn't cross line A-B.
if (line2.Y1 < 0 && line2.Y2 < 0 || line2.Y1 >= 0 && line2.Y2 >= 0)
return Intersection.None;
// (3) Discover the position of the intersection point along line A-B.
double posAB = line2.X2 + (line2.X1 - line2.X2) * line2.Y2 / (line2.Y2 - line2.Y1);
// Fail if segment C-D crosses line A-B outside of segment A-B.
if (posAB < 0 || posAB > distAB)
return Intersection.None;
// (4) Apply the discovered position to line A-B in the original coordinate system.
return Intersection.Intersection;
}
请注意，该方法会旋转线段以避免与方向相关的问题
答案 3 :(得分：1)
如果是2d，则所有行都在唯一的平面上。
所以，这是基本的三维几何。你应该可以用一个简单的方程式做到这一点。
查看此页面：
只要将矩形的坐标转换为平面方程，第二种解决方案就应该易于实现。
此外，检查您的分母是否为零（线不相交或包含在平面中）。
答案 4 :(得分：0)
使用课程：
System.Drawing.Rectangle
方法：
IntersectsWith();
答案 5 :(得分：-1)
我讨厌浏览MSDN文档（它们非常慢而且很奇怪：-s）但我认为它们应该有类似于this Java method的东西......如果它们没有，那么对它们不好！ XD（顺便说一句，它适用于细分，而不适用于细分）。
在任何情况下，您都可以查看开源Java SDK以了解它是如何实现的，也许您将学习一些新技巧（当我查看其他人的代码时，我总是感到惊讶）
答案 6 :(得分：-1)
是否可以使用简单的线段公式检查矩形每边的线。