Question

我有一个由PointF数组定义的GraphicsPath（由鼠标点击创建），由Graphics.DrawCurve绘制。然后转换路径以将起点和终点旋转到0度。转换和旋转的曲线（和原始曲线）PointF数组仍然可用。

我需要找到从路径的直线终点到曲线本身的最长垂直距离。

由于我已经旋转了曲线，找到最长的距离应该与边界的高度相同......但我还需要知道0轴的距离（现在是什么）（如果我能找到曲线那将很容易）。我已经进行了广泛的搜索 - 我尝试了自定义样条函数来尝试沿着曲线获得更多的固定点，但这最终以相同的结果结束 - 我不知道曲线是什么在点之间，并且在点之间插值的情况下可能是90％。

现在它已经旋转了，我已经尝试了一个for循环来计算顶部，并且在每一列中，试图查看每个点是否与变换路径相反的IsVisible，但它总是返回false

输入点（参考）：{X = 499，Y = 64} {X = 305，Y = 117} {X = 149，Y = 114}

（从左到右，我知道我需要做一些检查以确定将来它们的输入方式，但就目前而言，只是试图让它工作）

    GraphicsPath gP = new GraphicsPath();
gP.AddCurve(lPoints.ToArray());

Graphics g = pbImage.CreateGraphics();
g.SmoothingMode = SmoothingMode.AntiAlias;

//draws original curve (hiding for now to validate and visualize rotation:
//g.DrawPath(new Pen(Color.Blue, 1.75f), gP);
//g.DrawLine(new Pen(Color.Red, 1.75f), lPoints[0], lPoints[lPoints.Count - 1]);

// get angle in radians
double theta2 = Math.Atan2(lPoints[0].Y - lPoints[lPoints.Count - 1].Y, lPoints[0].X - lPoints[lPoints.Count - 1].X);

// convert radians to degrees
double angle = -1 * (theta2 * (180.0 / Math.PI));

//set a new path to the old one, to keep both sets of data
GraphicsPath newPath = gP;

//create rotational matrix, and transform
Matrix m = new Matrix();
m.RotateAt(Convert.ToSingle(angle), lPoints[0]);
newPath.Transform(m);

//draw transformed path to picture box
g.DrawPath(new Pen(Color.Green, 1f), newPath);

//get new points from transformed curve (interestingly enough, it adds more points than the oringial input)
PointF[] tP = newPath.PathPoints;

//create some temp variables to make the next section easier
PointF pS = tP[0];
PointF pE = tP[tP.Length - 1];

//draw the straight line for visual validation
g.DrawLine(new Pen(Color.Red, 1f), pS, pE);

//get the bounds of the new path
RectangleF bRect = newPath.GetBounds();
a
// loop through x's
for (float i = pE.X; i < pS.X; i = i + 5)
{
    float h = pS.Y - bRect.Y;
    bool bFound = false;

    // loop through y's - do loop until found
    do
    {
        if (newPath.IsVisible(i, h, g))
        {
            // never found
            tbOutPt.Text = tbOutPt.Text + "found!!!!";
            bFound = true;
        }
        h++;
    } while (bFound = false && h < bRect.Height);

}

我能想到的唯一另一件事就是根据边界创建一个新的位图，将曲线重绘为它，然后逐列逐行获取像素颜色，直到它与绘制曲线的颜色相匹配。

希望某人有更好的解决方案。

Answer 1

Flatten()使用GraphicsPath，然后使用标准的“点到线距离”测量值，遍历结果点并找到最大距离：

private float PointToLineDist(float Px, float Py, float Ax, float Ay, float Bx, float By)
{
    float q = 0;
    if ((Ax == Bx) & (Ay == By)) {
        // A and B passed in define a point, not a line.
        // Point to Point Distance
        return PointToPointDist(Px, Py, Ax, Ay);
    } else {
        // Distance is the length of the line needed to connect the point to
        // the(segment)such that the two lines would be perpendicular.

        // q is the parameterized value needed to get to the intersection
        q = ((Px - Ax) * (Bx - Ax) + (Py - Ay) * (By - Ay)) / ((Bx - Ax) * (Bx - Ax) + (By - Ay) * (By - Ay));

        // Limit q to 0 <= q <= 1
        // If q is outside this range then the Point is somewhere past the 
        // endpoints of our segment.  By setting q = 0 or q = 1 we are 
        // measuring the actual distacne from the point to one of the 
        // endpoints(instead)
        if (q < 0)
            q = 0;
        if (q > 1)
            q = 1;

        // Distance
        return PointToPointDist(Px, Py, (1 - q) * Ax + q * Bx, (1 - q) * Ay + q * By);
    }
}

private float PointToPointDist(float Ax, float Ay, float Bx, float By)
{
    // PointToPointDist = SquareRoot((Bx - Ax)^2 + (By - Ay)^2)
    return Math.Sqrt((Bx - Ax) * (Bx - Ax) + (By - Ay) * (By - Ay));
}

在PointToLineDist()中，（Px，Py）是有问题的点，（Ax，Ay），（Bx，By）是线段的端点。

Answer 2

以下是未来代码的结果代码：

private void processArray()
{

// this is for demo only - real points should come from mouse input, or otherwise
PointF[] inputArray = new PointF[2];

inputArray[0] = new PointF(537, 147);
inputArray[1] = new PointF(334, 180);
inputArray[2] = new PointF(150, 167);

GraphicsPath gP = new GraphicsPath();
gP.AddCurve(inputArray);

Graphics g = pbImage.CreateGraphics();
g.SmoothingMode = SmoothingMode.AntiAlias;

//draws original curve
g.DrawPath(new Pen(Color.Blue, 1.75f), gP);

// draw a straight line between the ends of the curve
//g.DrawLine(new Pen(Color.Red, 1.75f), lPoints[0], lPoints[lPoints.Count - 1]);

// create second path to flatten
GraphicsPath pathFlat = gP;
pathFlat.Flatten();

// get list of points to step through
PointF[] fP = pathFlat.PathPoints;

//variables to store max distance
float maxDistance = 0;
PointF maxDistP = new PointF(0, 0);

foreach (PointF p in fP)
{
    // get the distance from the point to the closet point on the line segment
    float curDist = PointToLineDist(p.X, p.Y, lPoints[0].X, lPoints[0].Y, lPoints[lPoints.Count - 1].X, lPoints[lPoints.Count - 1].Y);

    // check the value against and store the longest point
    if (curDist > maxDistance)
    {
        maxDistance = curDist;
        maxDistP = new PointF(p.X, p.Y);
    }
}

// mark a dot at the longest point
g.DrawRectangle(new Pen(Color.Red), new Rectangle(Convert.ToInt32(maxDistP.X), Convert.ToInt32(maxDistP.Y), 2, 2
}

这个功能应该全部包含在一个类中，这样你就可以创建和编辑图形路径并根据需要获取它们的属性。

C＃从直线到用户输入图形路径的最长距离

2 个答案: