偏移矩形位置以防止重叠

Question

我有以下矩形重叠的情况。

我有相交矩形（蓝色）的坐标。我如何偏移 2 个矩形的起始坐标（左上角 X-Y）以防止重叠，前提是我有使用

的交叉矩形的坐标

rectangle3 = Rectangle.Intersect(rectangle1, rectangle2);

更新：

根据@MBo 的更新，我添加了以下代码

Blue = Rectangle.Intersect(First, Second);
if (Blue != Rectangle.Empty)
{
if (First.Right == Blue.Right)
{

imgpoint.X += (Blue.Right - Blue.Left);
}

if (First.Bottom == Blue.Bottom)
{
imgpoint.Y += (Blue.Bottom - Blue.Top);
}

if (First.Left == Blue.Left)
{
imgpoint.X-= (Blue.Right - Blue.Left);
}
if (First.Top == Blue.Top)
{
imgpoint.Y -= (Blue.Bottom - Blue.Top);
}
}

更新2：

根据@MBo 的逻辑，我实现了代码 above.This 在某些图像中大多数 cases.But 有效，即使检测到相交并添加了偏移，我得到 overlapping.See下面的示例图片。

我试图解决的实际问题与这两个问题有关

Performing Overlap Detection of 2 Bitmaps

请指教。

更新：

更新：

根据您的更新，我已将 Delphi 代码翻译成 C#

Blue = Rectangle.Intersect(First, Second);
if (Blue != Rectangle.Empty)
{
int dcy,dcx;
int dy,dx;
int fl=First.Left,fw=First.Width,fh=First.Height,ft=First.Top,fb=First.Bottom,fr=First.Right;
int sl=Second.Left,sr=Second.Right,st=Second.Top,sb=Second.Bottom,sw=Second.Width,sh=Second.Height;
dcy = (sb + st) - (fb + ft);  //doubled center y-difference
dcx = (sl + sr) - (fl + fr);
if ((int)(dcx) >= (fw + sw) || ((int)(dcy) >= (fh + sh)))
{//no intersection

}
else
{

dx = fw + sw - (int)dcx;  //doubled  needed x-shift absolute value
dy = fh + sh - (int)dcy;

if (dx > dy)
{
if (dcy < 0)
{
dy = - dy/2;
}
else
{
dy = dy/ 2;  //needed y-shift accounting for direction
}
dx = 0;
}
else 
{
if (dcy < 0)
{
dx =  - dx/ 2;
}
else
{
dx =  dx/2;
dy = 0;
}
}
imgpoint.X+=dx;
imgpoint.Y+=dy;
}
}

使用此代码时，第二张图片在某些地方移动得太远 cases.Please 让我知道代码是否正确。

Answer 1

编辑： 检查了所有可能的情况，得到了下一个简单代码（Delphi 代码，有效）。

//fl, fw, fr, ft, fh, fb: First.Left, Width, Right, Top, Height, Bottom
//s* - similar parameters for the second rectangle   


dcy := (sb + st) - (fb + ft);  //doubled center y-difference
dcx := (sl + sr) - (fl + fr);
if (Abs(dcx) >= fw + sw) or ((Abs(dcy) >= fh + sh)) then  //no intersection
  Exit;

dx := fw + sw - Abs(dcx);  //doubled  needed x-shift absolute value
dy := fh + sh - Abs(dcy);

 if dx > dy then begin
   if dcy < 0 then
      dy := - dy div 2
   else
      dy := dy div 2;  //needed y-shift accounting for direction
   dx := 0;
 end else begin
   if dcy < 0 then
      dx :=  - dx div 2
   else
      dx :=  dx div 2;
   dy := 0;
 end;

//Result: dx, dy pair to shift the second rectangle

Full code

旧答案临时参考：

您需要信息 - 第一个矩形的坐标与相交矩形的相应坐标重合。对于简单的交集情况：

case
   First.Right = Blue.Right:   
      shift Second.Left by (Blue.Right - Blue.Left)
   First.Bottom = Blue.Bottom:
      shift Second.Top by (Blue.Bottom - Blue.Top)
   First.Left = Blue.Left:
      shift Second.Left by  -(Blue.Right - Blue.Left)
   First.Top = Blue.Top:
      shift Second.Top by -(Blue.Bottom - Blue.Top)

编辑：
如果存在交集并且满足上述 none 种情况，则发生完全包含和十字交集。所以确定避免重叠的最短方法是什么：

 dcy = (Second.Bottom + Second.Top) - (First.Bottom + First.Top)   
 if dcy >=0 then
      shift Second.Top by (First.Bottom - Second.Top)   //down
 else
      shift Second.Top by -(Second.Bottom - First.Top)   //up

 dcx = (Second.Left + Second.Right) - (First.Left + First.Right)   
 if dcx >=0 then
      shift Second.Left by (First.Right - Second.Left)   //right
 else
      shift Second.Left by -(Second.Right - First.Left)   //left

Answer 2

您需要检查 r2(rectangle2) 是否介于 r2.X 和 r2.X+r2.Height 之间或介于 r2.Y 和 r2.Y + r2.Y+r2.Width 或不在我们之间进行更改我写了一个示例代码

private int GetNewX(Rectangle r1, Rectangle r2)
{
   if (r2.X < r1.X)
   {
       return r1.X - r2.Width;
   }
   else
   {
       return r1.X + r1.Width;
   }
}
private int GetNewY(Rectangle r1, Rectangle r2)
{
   if (r2.Y < r1.Y)
   {
       return r1.Y - r2.Height;
   }
   else
   {
       return r1.Y + r1.Height;
   }
}

Pen pen = new Pen(Color.Black);
Rectangle r1 = new Rectangle(60, 10, 200, 200);
Rectangle r2 = new Rectangle(40, 25, 200, 160);


//If overlapped change X,Y of the r2 rectangle
Rectangle overlapRect = Rectangle.Intersect(r1, r2);
if (overlapRect.Width > 0 || overlapRect.Height > 0)
{
    bool betweenX = overlapRect.X >= r1.X && overlapRect.X <= (r1.X + r1.Height);
    bool betweenY = overlapRect.Y >= r1.Y && overlapRect.Y <= (r1.Y + r1.Width);

    if (betweenX)
    {
        r2.X = GetNewX(r1,r2);                    
    }
    else if (betweenY)
    {
        r2.Y = GetNewY(r1, r2);
    }
    else
    {
        if (overlapRect.Width <= overlapRect.Height)
        {
            r2.X = GetNewX(r1, r2);
        }
        else
        {
            r2.Y = GetNewY(r1, r2);
        }
    }                
}

Graphics g = this.CreateGraphics();
g.DrawRectangle(pen, r1);
g.DrawRectangle(pen, r2);

偏移矩形位置以防止重叠

Offsetting Rectangle Positions to prevent overlap

c#

bitmap

geometry

gdi+

system.drawing