求以知两点坐标画曲线的算法.

Question

求已知两点坐标画曲线的算法.
坐标A(x1,y1),B(x2,y2)
画一条平滑的曲线连结这两点的算法.谢谢.
C#做.

Answer 1

Pen CurvePan = new Pen(currentColor, 2);           if(p1.X>p1.X)
           {
               PointF ptem = new PointF();
               ptem=p1;
               p1=p2;
               p2=ptem;
           }            double  instance =Math.Sqrt( (p1.X - p2.X) * (p1.X - p2.X) - (p1.Y - p2.Y) * (p1.Y - p2.Y));            int height = Convert.ToInt16( Math.Abs(p1.Y - p2.Y));            int width=Convert.ToInt16( Math.Abs(p1.X - p2.X));            int total = height ;            PointF[] CurrP = new PointF[total];            double R = instance / 2.0f + height / 3.0f;
           
            for (int i = 0; i < total; i++)
            {
                float x = p1.X +(width/(total+.01f))*i;
                float y = p1.Y + (float)Math.Sqrt(R * R - x * x);
                PointF pch= new PointF(x, y);
                CurrP[i] = pch;
                         }
            g.DrawCurve(CurvePan, CurrP, 0.1f);
我的算法,结果不能连结两点

Answer 2

PointF[] CurrP = new PointF[2]{A(x1,y1),B(x2,y2) 
};  g.DrawCurve(CurvePan, CurrP); 
就可以了这是两点的
多的话就这样
List<PointF> list = new List<PointF>();
PointF pt0 = new PointF(x0,y0);
list.add(pt0);
PointF pt2 = new PointF(x1,y1);
list.add(pt1);
PointF pt2 = new PointF(x2,y2);
list.add(pt2);
PointF pt3 = new PointF(x3,y3);
list.add(pt3);
 PointF[] ptdfdf = new PointF[list.Count];
 list.CopyTo(ptdfdf);
 if (ptdfdf.Length > 1)
 {
      for (int i = 0; i < list.Count; i++)
      {
           g.DrawCurve(new Pen(cuvecolor), ptdfdf);
      }
 }

Answer 3

double  instance =Math.Sqrt( (p1.X - p2.X) * (p1.X - p2.X) - (p1.Y - p2.Y) * (p1.Y - p2.Y)); 
是什么意思啊？
听该是double  instance =Math.Sqrt( (p1.X - p2.X) * (p1.X - p2.X) + (p1.Y - p2.Y) * (p1.Y - p2.Y));吧