如何获取重叠的矩形坐标

呼叫 Rectangle.Intersect .

两个矩形的重叠区域的X坐标可以根据以下逻辑找到。

要找到y坐标,在四个假设中的最后一个,以及在这三个情况中,用y代替x。

假设:

• 一 和 乙 是矩形(其边沿x和y轴对齐),

• 每个矩形由两个点定义。( X _闽 / Y _闽 ) X _最大值 / Y _最大值 )

• 在哪里? X _闽 &; X _最大值 和 Y _闽 &; Y _最大值 .

• A.X _闽 &; B.X _闽

案例1-无重叠:

+--------+
|A       |    
|        |    +----+
|        |    |B   |
|        |    +----+
|        |
+--------+

A.X _闽 &; A.X _最大值 &; B.X _闽 &; B.X _最大值 无重叠。

案例2-部分重叠:

+--------+
|A       |
|     +--+-+
|     |B | |
|     +--+-+
|        |
+--------+

A.X _闽 &; B.X _闽 &; A.X _最大值 &; B.X _最大值 重叠X坐标: B.X _闽 - A.X _最大值

案例3-完全重叠:

+--------+
|A       |
| +----+ |
| |B   | |
| +----+ |
|        |
+--------+

A.X _闽 &; B.X _闽 &; B.X _最大值 &; A.X _最大值 重叠X坐标: B.X _闽 - B.X _最大值

附笔。: 实际上,您可以进一步简化这个算法。重叠X坐标始终是:

最大值(max) A.X _闽 , B.X _闽 -min( A.X _最大值 , B.X _最大值 )

除非第二个值小于第一个值;这意味着没有重叠。

static internal Rectangle intersect(Rectangle lhs, Rectangle rhs)
{
    Dimension lhsLeft = lhs.Location.X;
    Dimension rhsLeft = rhs.Location.X;
    Dimension lhsTop = lhs.Location.Y;
    Dimension rhsTop = rhs.Location.Y;
    Dimension lhsRight = lhs.Right;
    Dimension rhsRight = rhs.Right;
    Dimension lhsBottom = lhs.Bottom;
    Dimension rhsBottom = rhs.Bottom;

    Dimension left = Dimension.max(lhsLeft, rhsLeft);
    Dimension top = Dimension.max(lhsTop, rhsTop);
    Dimension right = Dimension.min(lhsRight, rhsRight);
    Dimension bottom = Dimension.min(lhsBottom, rhsBottom);
    Point location = new Point(left, top);
    Dimension width = (right > left) ? (right - left) : new Dimension(0);
    Dimension height = (bottom > top) ? (bottom - top) : new Dimension(0);

    return new Rectangle(location, new Size(width, height));
}

假设:

Points of   rectangle R1: R1.A(x,y), R1.B(x,y), R1.C(x,y), R1.D(x,y)   
Points of   rectangle R2: R2.A(x,y), R2.B(x,y), R2.C(x,y), R2.D(x,y)   
Overlapping rectangle RO: RO.A(x,y), RO.B(x,y), RO.C(x,y), RO.D(x,y)    
Standard cartesian coordinates (positive is right and upwards).

重叠矩形ro用c_计算如下:

RO.A.x = Math.Min(R1.A.x, R2.A.x);
RO.A.y = Math.Max(R1.A.y, R2.A.y);
RO.C.x = Math.Max(R1.C.x, R2.C.x);
RO.C.y = Math.Min(R1.C.y, R2.C.y);
RO.B(x,y) and RO.D(x,y) = ....

内矩形ri:

将上面解决方案中的最小值和最大值替换为重叠矩形ro。

我为我的项目使用了一个抽象验证器,并检查是否有一些布局控件在其中重叠,我在布局图中创建了矩形:

RuleFor(p => DoControlsIntersect(p.PageControls.Select(x => new Rectangle(x.Row, x.Column, x.Width, x.Height)).ToList())).Equal(false).WithMessage(OverlappingFields);

private bool DoControlsIntersect(List<Rectangle> rectangles)
        {
            return rectangles.Any(rect => rectangles.Where(r => !r.Equals(rect)).Any(r => r.IntersectsWith(rect)));
        }