计算机图形学常用算法实现7 Cohen-Sutherland裁剪算法

在winform下运行
算法本身的实现不算很难，但是这个算法的思路很秀，反正我是想不出来。
代码中的区域为(200,200)~(400,400)的区域

int encode(Point p)
{
    int code = 0;
    if (p.Y > 400)
        code += 8;
    if (p.Y < 200)
        code += 4;
    if (p.X > 400)
        code += 2;
    if (p.X < 200)
        code += 1;
    return code;
}
void lineClip(Point p1,Point p2)
{
    Graphics g = this.CreateGraphics();
    Pen p = new Pen(Brushes.Black);
    int code1 = encode(p1);
    int code2 = encode(p2);
    while (code1 != 0 || code2 != 0)
    {
        //在直线之外，直接舍去，但是跨过三个区域的情况舍弃不了，需要在后面分割一次再分别舍去
        if ((code1 & code2) != 0)
            return;
        //处理大于上界的点
        if ((code1 & 8) != 0)
        {
            p1.X -= (p1.Y - 400) * (p1.X - p2.X) / (p1.Y - p2.Y);
            p1.Y = 400;
            code1 = encode(p1);
        }
        else if ((code2 & 8) != 0)
        {
            p2.X -= (p2.Y - 400) * (p2.X - p1.X) / (p2.Y - p1.Y);
            p2.Y = 400;
            code2 = encode(p2);
        }
        //处理小于下界的点
        else if ((code1 & 4) != 0)
        {
            p1.X -= (p1.Y - 200) * (p1.X - p2.X) / (p1.Y - p2.Y);
            p1.Y = 200;
            code1 = encode(p1);
        }
        else if ((code2 & 4) != 0)
        {
            p2.X -= (p2.Y - 200) * (p2.X - p1.X) / (p2.Y - p1.Y);
            p2.Y = 200;
            code2 = encode(p2);
        }
        //处理大于右界的点
        else if ((code1 & 2) != 0)
        {
            p1.Y -= (p1.X - 400) * (p1.Y - p2.Y) / (p1.X - p2.X);
            p1.X = 400;
            code1 = encode(p1);
        }
        else if ((code2 & 2) != 0)
        {
            p2.Y -= (p2.X - 400) * (p2.Y - p1.Y) / (p2.X - p1.X);
            p2.X = 400;
            code2 = encode(p2);
        }
        //处理小于左边界的点
        else if ((code1 & 1) != 0)
        {
            p1.Y -= (p1.X - 200) * (p1.Y - p2.Y) / (p1.X - p2.X);
            p1.X = 200;
            code1 = encode(p1);
        }
        else if ((code2 & 1) != 0)
        {
            p2.Y -= (p2.X - 200) * (p2.Y - p1.Y) / (p2.X - p1.X);
            p2.X = 200;
            code2 = encode(p2);
        }
    }
    g.DrawLine(p, p1, p2);
}