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浮雕的算法:
下一行的红绿蓝颜色分量减去上一行的红绿蓝颜色分量,再加上128
如果小于0,那么等于0,如果大于255,那么等于255
R=R(i-1,j)-R(i,j)+128
G=G(i-1,j)-G(i,j)+128
B=B(i-1,j)-B(i,j)+128
主要代码如下:
'浮雕1
Private Sub btnEmboss_Click(sender As Object, e As EventArgs) Handles btnEmboss.Click
Dim pSourceColor1 As Color
Dim pSourceColor2 As Color
Dim pDestColor As Color
Dim destImg As New Bitmap(sourceImg.Width, sourceImg.Height)
Dim R, G, B As Integer
Dim R1, G1, B1 As Integer
Dim R2, G2, B2 As Integer
For i As Integer = 1 To sourceImg.Width - 1
For j As Integer = 0 To sourceImg.Height - 1
pSourceColor1 = sourceImg.GetPixel(i, j)
R1 = pSourceColor1.R
G1 = pSourceColor1.G
B1 = pSourceColor1.B
pSourceColor2 = sourceImg.GetPixel(i - 1, j)
R2 = pSourceColor2.R
G2 = pSourceColor2.G
B2 = pSourceColor2.B
R = R1 - R2 + 128
G = G1 - G2 + 128
B = B1 - B2 + 128
If R < 0 Then R = 0
If R > 255 Then R = 255
If G < 0 Then G = 0
If G > 255 Then G = 255
If B < 0 Then B = 0
If B > 255 Then B = 255
pDestColor = Color.FromArgb(R, G, B)
destImg.SetPixel(i, j, pDestColor)
Next
Next
picDest.Image = destImg
End Sub
运行效果如下:
我还看到了另外一个算法:
下一行的红绿蓝颜色分量减去上一行的红绿蓝颜色分量的绝对值,再加上128
如果小于0,那么等于0,如果大于255,那么等于255
R=Abs(R(i-1,j)-R(i,j))+128
G=Abs(G(i-1,j)-G(i,j))+128
B=Abs(B(i-1,j)-B(i,j))+128
主要代码如下:
'浮雕2
Private Sub btnEmboss2_Click(sender As Object, e As EventArgs) Handles btnEmboss2.Click
Dim pSourceColor1 As Color
Dim pSourceColor2 As Color
Dim pDestColor As Color
Dim destImg As New Bitmap(sourceImg.Width, sourceImg.Height)
Dim R, G, B As Integer
Dim R1, G1, B1 As Integer
Dim R2, G2, B2 As Integer
For i As Integer = 1 To sourceImg.Width - 1
For j As Integer = 0 To sourceImg.Height - 1
pSourceColor1 = sourceImg.GetPixel(i, j)
R1 = pSourceColor1.R
G1 = pSourceColor1.G
B1 = pSourceColor1.B
pSourceColor2 = sourceImg.GetPixel(i - 1, j)
R2 = pSourceColor2.R
G2 = pSourceColor2.G
B2 = pSourceColor2.B
R = Math.Min(Math.Abs(R1 - R2) + 128, 255)
G = Math.Min(Math.Abs(G1 - G2) + 128, 255)
B = Math.Min(Math.Abs(B1 - B2) + 128, 255)
pDestColor = Color.FromArgb(R, G, B)
destImg.SetPixel(i, j, pDestColor)
Next
Next
picDest.Image = destImg
End Sub
其中Math.Min(A,B)将返回A和B中较小的值。
运行效果如下:
由于.net平台下C#和vb.NET很相似,本文也可以为C#爱好者提供参考。
学习更多vb.net知识,请参看 vb.net教程目录
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