delphi - 如何加快算法(二值化，积分图像)

Question

我根据此网址中描述的积分图像算法编写了此程序

http://people.scs.carleton.ca/~roth/iit-publications-iti/docs/gerh-50002.pdf

有什么办法可以更快地执行此代码？

指针比动态数组快得多吗？

procedure TForm1.bBinarizationClick(Sender: TObject);
var
  iX1, iY1,
  iX2, iY2,
  ii, jj,
  s, s2,
  iSum, iCount,  index,
  iHeight, iWidth : Integer;
  iSize: Integer;

  row : ^TRGBTriple;
  black : TRGBTriple;
  aIntegralIm: array  of Integer;
  aGrays : array of Byte;

  startTime : Cardinal;

begin
  iWidth := bBitmap.Width;
  iHeight := bBitmap.Height;
  iSize := iWidth * iHeight;

  SetLength(aGrays, iSize);
  SetLength(aIntegralIm, iSize);

  black.rgbtRed  := (clBlack and $0000FF);
  black.rgbtGreen := (clBlack and $00FF00) shr 8;
  black.rgbtBlue := (clBlack and $FF0000) shr 16;

  bBitmap2.Canvas.Brush.Color := clWhite;
  bBitmap2.Canvas.FillRect(Rect(0, 0, bBitmap2.Width, bBitmap2.Height));

  s := Round(iWidth / TrackBar2.Position);
    s2 := Round(s / 2);

  startTime := GetTickCount();

  index := 0;

  for ii := 0 to iHeight - 1 do begin
     row := bBitmap.ScanLine[ii];
     for jj := 0 to iWidth - 1 do begin
       aGrays[index] := ((row.rgbtRed * 77 + row.rgbtGreen * 150 + row.rgbtBlue * 29) shr 8);
       inc(index);
       inc(row);
     end;
  end;


  for ii := 0 to iWidth - 1 do begin
     iSum := 0;
     for jj := 0 to iHeight - 1 do begin
       index := jj*iWidth+ii;
       iSum := iSum + aGrays[index];
       if ii = 0 then aIntegralIm[index] := iSum
       else aIntegralIm[index] := aIntegralIm[index - 1] + iSum;
     end;
  end;


  for jj := 0 to iHeight - 1 do begin
     row := bBitmap2.ScanLine[jj];
     for ii := 0 to iWidth - 1 do begin

       index := jj*iWidth+ii;

       iX1 := ii-s2;
       iX2 := ii+s2;
       iY1 := jj-s2;
       iY2 := jj+s2;

       if (iX1 < 0) then iX1 := 0;
         if (iX2 >= iWidth) then  iX2 := iWidth-1;
           if (iY1 < 0) then  iY1 := 0;
             if (iY2 >= iHeight) then  iY2 := iHeight-1;

       iCount := (iX2 - iX1) * (iY2 - iY1);

       iSum := aIntegralIm[iY2*iWidth+iX2]
              - aIntegralIm[iY1*iWidth+iX2]
              - aIntegralIm[iY2*iWidth+iX1]
              + aIntegralIm[iY1*iWidth+iX1];

       if (aGrays[index] * iCount) < (iSum * (100 - TrackBar1.Position) / 100) then  row^ :=  black;

       inc(row);

     end;
  end;

  ePath.Text :=  'Time: ' + inttostr(GetTickCount() - startTime) + ' ms';

  imgOryginal.Picture.Bitmap.Assign(bBitmap2);

end;

Answer 1

您至少可以做一些简单的事情：


将（100-TrackBar1.Position）预计算为变量
代替除法：/ 100在另一侧使用* 100。您可能不需要任何浮点值。
使用查找表进行以下操作（请注意解释标识btw？）：


码：

if (iX1 < 0) then iX1 := 0;
if (iX2 >= iWidth) then  iX2 := iWidth-1;
if (iY1 < 0) then  iY1 := 0;
if (iY2 >= iHeight) then  iY2 := iHeight-1;

尝试保持索引和icremnet减量而不是乘法：index：= jj * iWidth + ii;