实施Graham Scan以找到凸包

Question

我正在尝试用C++ 实现Graham Scan,但它不起作用,我找不到原因.任何领导将不胜感激.经过一些尝试后,似乎我总是有m_M = 22点是最高点,如果有帮助的话.

交叉产品,知道它是右转还是左转.

qreal Interpolation::ccw(QPointF pt1, QPointF pt2, QPointF pt3)
{
    return (pt2.x()-pt1.x())*(pt3.y()-pt1.y()) - (pt2.y()-pt1.y())*(pt3.x()-pt1.x());
}

点积除以常数,cos因为排序角度与排序相同cos in [0, Pi].

qreal Interpolation::dp(QPointF pt1, QPointF pt2)
{
    return (pt2.x()-pt1.x())/qSqrt((pt2.x()-pt1.x())*(pt2.x()-pt1.x()) + (pt2.y()-pt1.y())*(pt2.y()-pt1.y()));
}

主要功能:

void Interpolation::ConvexHull()
{
    QPointF points[m_N+1]; // My number of points
    QPointF pt_temp(m_pt[0]);
    qreal angle_temp(0);
    int k_temp(0);

使用points[1]低y点填充数组点:

    for (int i(1); i < m_N; ++i)
    {
        if (m_pt[i].y() < pt_temp.y())
        {
            points[i+1] = pt_temp;
            pt_temp = m_pt[i];
        }
        else
        {
            points[i+1] = m_pt[i];
        }
    }
    points[1] = pt_temp;

按角度对点阵进行排序 points[m_N] = points[0]

    for (int i(2); i <= m_N; ++i)
    {
        pt_temp = points[i];
        angle_temp = dp(points[1], pt_temp);
        k_temp = i;
        for (int j(1); j <= m_N-i; ++j)
        {
            if (dp(points[1], points[i+j]) < angle_temp)
            {
                pt_temp = points[i+j];
                angle_temp = dp(points[1], points[i+j]);
                k_temp = i+j;
            }
        }
        points[k_temp] = points[i];
        points[i] = pt_temp;
    }
    points[0] = points[m_N];

执行格雷厄姆扫描

    m_M = 1; // Number of points on the convex hull.

    for (int i(2); i <= m_N; ++i)
    {
        while (ccw(points[m_M-1], points[m_M], points[i]) <= 0)
        {
            if (m_M > 1)
            {
                m_M -= 1;
            }
            else if (i == m_N)
            {
                break;
            }
            else
            {
                i += 1;
            }
        }
        m_M += 1;
        pt_temp = points[m_M];
        points[m_M] = points[i];
        points[i] = points[m_M];
    }

保存m_convexHull应该是船体上的点列表的点ConvexHull[m_M]=[ConvexHull[0]

    for (int i(0); i < m_M; ++i)
    {
        m_convexHull.push_back(points[i+1]);
    }
    m_convexHull.push_back(points[1]);
}

Answer 1

我发现了这个问题.它的句子在于:

点积除以范数得到cos,因为对角度进行排序与​​在[0,Pi]中对cos进行排序相同.

角度越低,cos越高,所以我只需更改这行代码:

            if (dp(points[1], points[i+j]) < angle_temp)

至:

            if (dp(points[1], points[i+j]) > angle_temp)

现在它完美无缺!