计算三次贝塞尔曲线的最快方法？

Question

现在我是这样计算的：

    double dx1 = a.RightHandle.x - a.UserPoint.x;
    double dy1 = a.RightHandle.y - a.UserPoint.y;
    double dx2 = b.LeftHandle.x - a.RightHandle.x;
    double dy2 = b.LeftHandle.y - a.RightHandle.y;
    double dx3 = b.UserPoint.x - b.LeftHandle.x;
    double dy3 = b.UserPoint.y - b.LeftHandle.y;

    float len = sqrt(dx1 * dx1 + dy1 * dy1) + 
        sqrt(dx2 * dx2 + dy2 * dy2) + 
        sqrt(dx3 * dx3 + dy3 * dy3);




    int NUM_STEPS =  int(len * 0.05);

    if(NUM_STEPS > 55)
    {
        NUM_STEPS = 55;
    }
    double subdiv_step  = 1.0 / (NUM_STEPS + 1);
    double subdiv_step2 = subdiv_step*subdiv_step;
    double subdiv_step3 = subdiv_step*subdiv_step*subdiv_step;

    double pre1 = 3.0 * subdiv_step;
    double pre2 = 3.0 * subdiv_step2;
    double pre4 = 6.0 * subdiv_step2;
    double pre5 = 6.0 * subdiv_step3;



    double tmp1x = a.UserPoint.x - a.RightHandle.x * 2.0 + b.LeftHandle.x;
    double tmp1y = a.UserPoint.y - a.RightHandle.y  * 2.0 + b.LeftHandle.y;

    double tmp2x = (a.RightHandle.x - b.LeftHandle.x)*3.0 - a.UserPoint.x + b.UserPoint.x;
    double tmp2y = (a.RightHandle.y - b.LeftHandle.y)*3.0 - a.UserPoint.y + b.UserPoint.y;

    double fx = a.UserPoint.x;
    double fy = a.UserPoint.y;

    //a user
    //a right
    //b left
    //b user

    double dfx = (a.RightHandle.x - a.UserPoint.x)*pre1 + tmp1x*pre2 + tmp2x*subdiv_step3;
    double dfy = (a.RightHandle.y - a.UserPoint.y)*pre1 + tmp1y*pre2 + tmp2y*subdiv_step3;

    double ddfx = tmp1x*pre4 + tmp2x*pre5;
    double ddfy = tmp1y*pre4 + tmp2y*pre5;

    double dddfx = tmp2x*pre5;
    double dddfy = tmp2y*pre5;

    int step = NUM_STEPS;



    while(step--)
    {


        fx   += dfx;
        fy   += dfy;
        dfx  += ddfx;
        dfy  += ddfy;
        ddfx += dddfx;
        ddfy += dddfy;
        temp[0] = fx;
        temp[1] = fy;
        Contour[currentcontour].DrawingPoints.push_back(temp);
    }


    temp[0] = (GLdouble)b.UserPoint.x;
    temp[1] = (GLdouble)b.UserPoint.y;
    Contour[currentcontour].DrawingPoints.push_back(temp);

我想知道是否有更快的方法来插值三次贝塞尔曲线？

谢谢

Answer 1

研究前向差分以获得更快的方法。必须小心处理舍入误差。

自适应细分方法，通过一些检查，可以快速且准确。