優化GDI +功能的性能

在分析我的GDI +項目時，我發現以下IsLineVisible函數是在我的自定義面板上繪製和移動對象時「最熱」的函數之一。優化GDI +功能的性能

有沒有可能優化它？

Private Function IsLineVisible(ByVal detectorRectangle As Rectangle, 
           ByVal pen As Pen, 
           ByVal ParamArray points() As Point) As Boolean 
    Using path As New GraphicsPath() 
     path.AddLines(points) 
     Return IsPathVisible(detectorRectangle, path, pen) 
    End Using 
    End Function 

    ' Helper functions ''''''''''''''''''''''''''''''''''''' 
    Private Function IsPathVisible(ByVal detectorRectangle As Rectangle, 
           ByVal path As GraphicsPath, 
           ByVal pen As Pen) As Boolean 
    If Not path.IsPoint Then 
     path.Widen(pen) 
    End If 
    Return IsPathVisible(detectorRectangle, path) 
    End Function 


    Private Function IsPathVisible(ByVal detectorRectangle As Rectangle, 
           ByVal path As GraphicsPath) As Boolean 
    Using r As New Region(path) 
     If r.IsVisible(detectorRectangle) Then 
     Return True 
     Else 
     Return False 
     End If 
    End Using 
    End Function

enter image description here

來源

2011-10-11 serhio

更新2：

public bool AreLinesVisible(Point[] p, int width, Rectangle rect) 
    { 
     for (var i = 1; i < p.Length; i++) 
      if (IsLineVisible(p[i - 1], p[i], width, rect)) 
       return true; 
     return false; 
    }

修訂包括厚度/寬度。

這是完全未經測試的代碼，但它應該給你基本思路對於沒有昂貴的framwork一個超快速的解決方案要求：

public bool IsLineVisible(Point p1, Point p2, int width, Rectangle rect) 
{ 
    var a = Math.Atan2(p1.Y - p2.Y, p1.X - p2.X) + Math.PI/2; 
    var whalf = (width + 1)*0.5; 
    var dx = (int) Math.Round(whalf*Math.Sin(a)); 
    var dy = (int) Math.Round(whalf*Math.Cos(a)); 
    return IsLineVisible(new Point(p1.X - dx, p1.Y - dy), new Point(p2.X - dx, p2.Y - dy), rect) 
     || IsLineVisible(new Point(p1.X + dx, p1.Y + dy), new Point(p2.X + dx, p2.Y + dy), rect); 
} 

public bool IsLineVisible(Point p1, Point p2, Rectangle rect) 
{ 
    if (p1.X > p2.X) // make sure p1 is the leftmost point 
     return IsLineVisible(p2, p1, rect); 

    if (rect.Contains(p1) || rect.Contains(p2)) 
     return true; // one or both end-points within the rect -> line is visible 

    //if both points are simultaneously left or right or above or below -> line is NOT visible 
    if (p1.X < rect.X && p2.X < rect.X) 
     return false; 
    if (p1.X >= rect.Right && p2.X >= rect.Right) 
     return false; 
    if (p1.Y < rect.Y && p2.Y < rect.Y) 
     return false; 
    if (p1.Y >= rect.Bottom && p2.Y >= rect.Bottom) 
     return false; 

    // now recursivley break down the line in two part and see what happens 
    // (this is an approximation...) 
    var pMiddle = new Point((p1.X + p2.X)/2, (p1.Y + p2.Y)/2); 
    return IsLineVisible(p1, new Point(pMiddle.X - 1, pMiddle.Y), rect) 
      || IsLineVisible(new Point(pMiddle.X + 1, pMiddle.Y), p2, rect); 
}

來源

2011-10-11 17:09:06

謝謝。但是你「忘記」了筆的寬度**（厚度）。 – serhio

是的，但我說的基本思想 - 你只需要計算兩條輪廓線，組成你的粗線並通過它們。如果其中一個是可見的 - 那麼你的粗線是可見的！ –

這條線也可以有一個樣式：虛線，虛線... – serhio