遍历2.5D网格

Question

我正试图找出如何以有效的方式遍历2.5D网格。栅格本身是二维的，但栅格中的每个单元都有一个最小/最大浮动高度。要遍历的线由两个三维浮点坐标定义。如果进入/退出网格单元之间的z值范围与该单元的最小/最大高度不重叠，我希望停止遍历该线。

我目前正在使用2D DDA算法按顺序遍历网格单元格（见图），但我不确定如何在到达每个网格单元格时计算z值。如果可以的话，我可以在进入/离开单元格时根据单元格的最小/最大高度测试z值。

有没有办法修改这个算法，允许在输入每个网格单元格时计算z？或者有没有更好的遍历算法允许我这样做？

以下是我当前使用的代码：

void Grid::TraceGrid(Point3<float>& const start, Point3<float>& const end, GridCallback callback )
{
    // calculate and normalize the 2D direction vector
    Point2<float> direction=end-start;
    float length=direction.getLength( );
    direction/=length;

    // calculate delta using the grid resolution
    Point2<float> delta(m_gridresolution/fabs(direction.x), m_gridresolution/fabs(direction.y));

    // calculate the starting/ending points in the grid
    Point2<int> startGrid((int)(start.x/m_gridresolution), (int)(start.y/m_gridresolution));
    Point2<int> endGrid((int)(end.x/m_gridresolution), (int)(end.y/m_gridresolution));
    Point2<int> currentGrid=startGrid;

    // calculate the direction step in the grid based on the direction vector
    Point2<int> step(direction.x>=0?1:-1, direction.y>=0?1:-1);

    // calculate the distance to the next grid cell from the start
    Point2<float> currentDistance(((step.x>0?start.x:start.x+1)*m_gridresolution-start.x)/direction.x, ((step.y>0?start.y:start.y+1)*m_gridresolution-start.y)/direction.y);

    while(true)
    {
        // pass currentGrid to the callback
        float z = 0.0f;     // need to calculate z value somehow
        bool bstop=callback(currentGrid, z);

        // check if the callback wants to stop or the end grid cell was reached
        if(bstop||currentGrid==endGrid) break;

        // traverse to the next grid cell
        if(currentDistance.x<currentDistance.y) {
            currentDistance.x+=delta.x;
            currentGrid.x+=step.x;
        } else {
            currentDistance.y+=delta.y;
            currentGrid.y+=step.y;
        }
    }
}

Answer 1

我想出了一个好办法。在函数开头添加：

float fzoffset=end.z-start.z;
Point2<float> deltaZ(fzoffset/fabs(end.x-start.x), fzoffset/fabs(end.y-start.y));
Point2<float> currentOffset((step.x>0?start.x:start.x+1)*m_gridresolution-start.x, (step.y>0?start.y:start.y+1)*m_gridresolution-start.y);

在currentDistance所在的回路内。x/。y递增，添加：

currentOffset.x+=m_gridresolution;  //When stepping in the x axis
currentOffset.y+=m_gridresolution;  //When stepping in the y axis

然后在每一步计算z：

z=currentOffset.x*deltaZ.x+start.z;  //When stepping in the x axis
z=currentOffset.y*deltaZ.y+start.z;  //When stepping in the y axis

Answer 2

对于每个单元格，你知道你来自哪个单元格。这意味着你知道你来自哪一边。在绿线和给定网格线的交叉点计算z似乎微不足道。

Answer 3

Bresenham线算法的3D扩展似乎可以奏效。您将迭代X并独立跟踪线段的Y和Z分量的错误，以确定每个对应X值的Y和Z值。当Z中的累积误差达到某个临界水平时，您只需停止，这表明它超出了最小/最大值。