请教DrawCurve曲线上点坐标的方法？

xnlm2005 2012-04-23 11:37:12

或者别的方法，重要的是，平滑和获取曲线上的没有赋值的点坐标。

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crackdung 2012-04-27

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寫這麼多，不如用控件省事

LightningChart Ultimate 4.1

xnlm2005 2012-04-27

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找到了一个三次插值方法，求高手翻译一段程序，就当这个帖子的问题吧：



/**************************

** 三次样条插值函数.C

** 进行自然边界，

**     夹持边界，

**     非扭结边界

** 条件下的插值

** 2008-12-27

**************************/

#include <stdio.h>

#include <math.h>



typedef enum _condition

{

    NATURAL,

    CLAMPED,

    NOTaKNOT

}condition;



typedef struct _coefficient

{

    double a3;

    double b2;

    double c1;

    double d0;

}coefficient;



typedef struct _point

{

    coefficient *coe;

    double *xCoordinate;

    double *yCoordinate;

    double f0;

    double fn;

    int num;

    condition con;

}point;







int spline(point *point)

{

    double *x = (*point).xCoordinate, *y = (*point).yCoordinate;

    int n = (*point).num - 1;

    coefficient *coe = (*point).coe;

    condition con = (*point).con;

    double *h, *d;

    double *a, *b, *c, *f, *m;

    double temp;

    int i;

    h = (double *)malloc( n * sizeof(double) ); /* 0,1--(n-1),n */

    d = (double *)malloc( n * sizeof(double) ); /* 0,1--(n-1),n */

    

    a = (double *)malloc( (n + 1) * sizeof(double) );  /* 特别使用,1--(n-1),       n */

    b = (double *)malloc( (n + 1) * sizeof(double) );  /*        0,1--(n-1),       n */

    c = (double *)malloc( (n + 1) * sizeof(double) );  /*        0,1--(n-1),特别使用 */

    f = (double *)malloc( (n + 1) * sizeof(double) );  /*        0,1--(n-1),       n */

    m = b;

    if(f == NULL)

    {

        free(h);

        free(d);

        free(a);

        free(b);

        free(c);

        return 1;

    }

    /* 计算 h[] d[] */

    for (i = 0; i < n; i++)

    {

        h[i] = x[i + 1] - x[i];

        d[i] = (y[i + 1] - y[i]) / h[i];

        /* printf("%f\t%f\n", h[i], d[i]); */

    }

    /**********************

    ** 初始化系数增广矩阵

    **********************/

    a[0] = (*point).f0;

    c[n] = (*point).fn;

    /* 计算 a[] b[] d[] f[] */

    switch(con)

    {

        case NATURAL:

            f[0] = a[0];

            f[n] = c[n];

            a[0] = 0;

            c[n] = 0;

            c[0] = 0;

            a[n] = 0;

            b[0] = 1;

            b[n] = 1;

        break;

        

        case CLAMPED:

            f[0] = 6 * (d[0] - a[0]);

            f[n] = 6 * (c[n] - d[n - 1]);

            a[0] = 0;

            c[n] = 0;

            c[0] = h[0];

            a[n] = h[n - 1];

            b[0] = 2 * h[0];

            b[n] = 2 * h[n - 1];

        break;

        

        case NOTaKNOT:

            f[0] = 0;

            f[n] = 0;

            a[0] = h[0];

            c[n] = h[n - 1];

            c[0] = -(h[0] + h[1]);

            a[n] = -(h[n - 2] + h[n - 1]);

            b[0] = h[1];

            b[n] = h[n - 2];

        break;

    }



    for (i = 1; i < n; i++)

    {

        a[i] = h[i - 1];

        b[i] = 2 * (h[i - 1] + h[i]);

        c[i] = h[i];

        f[i] = 6 * (d[i] - d[i - 1]);

    }

    /* for (i = 0; i < n+1; i++)   printf("%f\n", c[i]); */

    

    /*************

    ** 高斯消元

    *************/

    /* 第0行到第(n-3)行的消元 */

    for(i = 0; i <= n - 3; i++)

    {

        /* 选主元 */

        if( fabs(a[i + 1]) > fabs(b[i]) )

        {

            temp = a[i + 1]; a[i + 1] = b[i]; b[i] = temp;

            temp = b[i + 1]; b[i + 1] = c[i]; c[i] = temp;

            temp = c[i + 1]; c[i + 1] = a[i]; a[i] = temp;

            temp = f[i + 1]; f[i + 1] = f[i]; f[i] = temp;

        }

        temp = a[i + 1] / b[i];

        a[i + 1] = 0;

        b[i + 1] = b[i + 1] - temp * c[i];

        c[i + 1] = c[i + 1] - temp * a[i];

        f[i + 1] = f[i + 1] - temp * f[i];

    }

    

    /* 最后3行的消元 */

    /* 选主元 */

    if( fabs(a[n - 1]) > fabs(b[n - 2]) )

    {

        temp = a[n - 1]; a[n - 1] = b[n - 2]; b[n - 2] = temp;

        temp = b[n - 1]; b[n - 1] = c[n - 2]; c[n - 2] = temp;

        temp = c[n - 1]; c[n - 1] = a[n - 2]; a[n - 2] = temp;

        temp = f[n - 1]; f[n - 1] = f[n - 2]; f[n - 2] = temp;

    }

    /* 选主元 */

    if( fabs(c[n]) > fabs(b[n - 2]) )

    {

        temp = c[n]; c[n] = b[n - 2]; b[n - 2] = temp;

        temp = a[n]; a[n] = c[n - 2]; c[n - 2] = temp;

        temp = b[n]; b[n] = a[n - 2]; a[n - 2] = temp;

        temp = f[n]; f[n] = f[n - 2]; f[n - 2] = temp;

    }

    /* 第(n-1)行消元 */

    temp = a[n - 1] / b[n - 2];

    a[n - 1] = 0;

    b[n - 1] = b[n - 1] - temp * c[n - 2];

    c[n - 1] = c[n - 1] - temp * a[n - 2];

    f[n - 1] = f[n - 1] - temp * f[n - 2];

    /* 第n行消元 */

    temp = c[n] / b[n - 2];

    c[n] = 0;

    a[n] = a[n] - temp * c[n - 2];

    b[n] = b[n] - temp * a[n - 2];

    f[n] = f[n] - temp * f[n - 2];

    /* 选主元 */

    if( fabs(a[n]) > fabs(b[n - 1]) )

    {

        temp = a[n]; a[n] = b[n - 1]; b[n - 1] = temp;

        temp = b[n]; b[n] = c[n - 1]; c[n - 1] = temp;

        temp = f[n]; f[n] = f[n - 1]; f[n - 1] = temp;

    }

    /* 最后一次消元 */

    temp = a[n] / b[n-1];

    a[n] = 0;

    b[n] = b[n] - temp * c[n - 1];

    f[n] = f[n] - temp * f[n - 1];

    

    /* 回代求解 m[] */

    m[n] = f[n] / b[n];

    m[n - 1] = (f[n - 1] - c[n - 1] * m[n]) / b[n-1];

    for(i = n - 2; i >= 0; i--)

    {

        m[i] = ( f[i] - (m[i + 2] * a[i] + m[i + 1] * c[i]) ) / b[i];

    }

    /* for (i = 0; i < n+1; i++)   printf("%f\n", m[i]); */

    free(a);

    free(c);

    free(f);

    /************

    ** 放置参数

    ************/

    for(i = 0; i < n; i++)

    {

        coe[i].a3 = (m[i + 1] - m[i]) / (6 * h[i]);

        coe[i].b2 = m[i] / 2;

        coe[i].c1 = d[i] - (h[i] / 6) * (2 * m[i] + m[i + 1]);

        coe[i].d0 = y[i];

    }

    

    free(h);

    free(d);

    free(b);

    return n + 1;

}



void main()

{

    /* x[]内不能出现重复数据 */

    double x[] = { 0, 0.5, 0.75, 1.25, 2.5,  5, 7.5, 10, 15,

                  20,  25,   30,   35,  40, 45,  50, 55, 60,

                  65,  70,   75,   80,  85, 90,  95, 100};

    double y[] = {     0, 0.6107, 0.7424, 0.9470, 1.3074, 1.7773,    2.1,

                  2.3414, 2.6726, 2.8688, 2.9706, 3.0009, 2.9743, 2.9015,

                  2.7904, 2.6470, 2.4762, 2.2817, 2.0662, 1.8320, 1.5802,

                  1.3116, 1.0263, 0.7239, 0.4033, 0.0630};

    int n = sizeof(x) / sizeof(double);

    coefficient *coe;

    int i;

    point p;

    coe = (coefficient *)malloc((n - 1) * sizeof(coefficient));

    p.xCoordinate = x;

    p.yCoordinate = y;

    p.f0 = 0;

    p.fn = 0;

    p.num = n;

    p.con = NOTaKNOT;

    p.coe = coe;

    spline(&p);

    

    for(i = 0; i < n - 1; i++)

        printf("%f\t%f\t%f\t%f\n", coe[i].a3, coe[i].b2, coe[i].c1, coe[i].d0);

    free(coe);

}

xnlm2005 2012-04-26

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[Quote=引用 7 楼的回复:]
System.Drawing.Drawing2D.GraphicsPath.AddCurve方法和IsOutlineVisible方法。前者添加曲线至路径，后者用于测试点是否在曲线上
[/Quote]
这样只能判断已知点，不能判断未知点。

xnlm2005 2012-04-26

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[Quote=引用 3 楼的回复:]
你怎么画的曲线呢，肯定有点的坐标数组，里面查去
[/Quote]
由于我用了很少的点生成了曲线，我现在需要在曲线上读取一些没有设置的点。

vinjent 2012-04-24

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private void pictureBox1_Paint(object sender, PaintEventArgs e)
{
e.Graphics.TranslateTransform(0, pictureBox1.Height / 2);
Pen pen = new Pen(Color.White, 2);

point[0] = new Point(0, 10);
point[1] = new Point(50, 20);
point[2] = new Point(100, -40);
point[3] = new Point(150, 10);
point[4] = new Point(200, -50);

e.Graphics.DrawCurve(pen, point);

}

private void pictureBox1_MouseMove(object sender, MouseEventArgs e)
{
for ( int i = 0; i < point.Length; i++)
{
if (point[i].X == e.X && point[i].Y == e.Y)
{
label1.Text = e.X.ToString();
label2.Text = e.Y.ToString();
}
}
}

WAN 2012-04-24

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System.Drawing.Drawing2D.GraphicsPath.AddCurve方法和IsOutlineVisible方法。前者添加曲线至路径，后者用于测试点是否在曲线上

xxgclj 2012-04-24

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试试DrawBezier

assky124 2012-04-24

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已知曲线图形，计算控制点？

bdmh 2012-04-24

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你怎么画的曲线呢，肯定有点的坐标数组，里面查去

xnlm2005 2012-04-24

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[Quote=引用 1 楼的回复:]

private void pictureBox1_Paint(object sender, PaintEventArgs e)
{
e.Graphics.TranslateTransform(0, pictureBox1.Height / 2);
Pen pen = new Pen(Color.White, 2);

……
[/Quote]

我不是判断鼠标是否在线上，并且你的方法肯定有问题，比如我点三个距离很远点，画一条曲线，你这样肯定判断不出来的，我的要求是获取在这三个指定点中间并且在这条曲线上的点的坐标并存为文件。