我有一个1平方公里的正方形，我把它分为正负两侧100米..
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基本上我正在制作一个像上面那样的正方形。我在x轴的正侧和负侧都潜入100米的正方形。然后我在他们各自的合作中绘制每个点，我打算通过一堆if / else语句（至少50，最多100）抛出（x，y）值，但我关心的是这将是昂贵的。有没有更有效的方式来做我想做的事情？
以下是我计划如何做的一个例子......
if(tlat < lat && tlat > lat - 89930.7){
//point is within 100 meters on lat
//On the NEGATIVE SIDE.
if(tlng > lng && tlng < lng + 147999.8){
//point is within 100 meters on lat NEGATIVE &&
//withing 100 meters on lng POSITIVE
layers.addcube(highlng100p, lowlng100p, yhigh, ylow, highlat100p, lowlat100p);
highlng100p = highlng100p + 5;
lowlng100p = lowlng100p + 5;
highlat100p = highlat100p + 5;
lowlat100p = lowlat100p + 5;
}else if(tlng < lng && tlng > lng - 147999.8){
//point is within 100 meters on lat NEGATIVE &&
//withing 100 meters on lng NEGATIVE
layers.addcube(highlat100n, lowlat100n, yhigh, ylow, highlng100n, lowlng100n);
highlat100n = highlat100n + 5;
lowlat100n = lowlat100n + 5;
highlng100n = highlng100n + 5;
lowlat100n = lowlat100n + 5;
}else if(tlng > lng && tlng < lng + 295999.6){
//point is within 200 meters on lat NEGATIVE &&
//withing 200 meters on lng POSITIVE
layers.addcube(highlat200n, lowlat200n, yhigh, ylow, highlng200p, lowlng200n);
highlat200n = highlat200n + 5;
lowlat200n = lowlat200n + 5;
highlng200p = highlng200p + 5;
lowlng200p = lowlng200p + 5;
}else if(tlng < lng && tlng > lng - 295999.6){
//point is within 200 meters on lat NEGATIVE &&
//withing 200 meters on lng NEGATIVE
layers.addcube(highlat200n, lowlat200n, yhigh, ylow, highlng200n, lowlng200n);
highlat200n = highlat200n + 5;
lowlat200n = lowlat200n + 5;
highlng200n = highlng200n + 5;
lowlng200n = lowlng200n + 5;
}else if(tlng > lng && tlng < lng + 443999.4){
//point is within 300 meters on lat NEGATIVE &&
//withing 300 meters on lng POSITIVE
layers.addcube(highlat300n, lowlat300n, yhigh, ylow, highlng300p, lowlng300p);
highlat300n = highlat300n + 5;
lowlat300n = lowlat300n + 5;
highlng300p = highlng300p + 5;
lowlng300p = lowlng300p + 5;
}else if(tlng < lng && tlng > lng - 443999.4){
//point is within 300 meters on lat NEGATIVE &&
//withing 300 meters on lng NEGATIVE
layers.addcube(highlat300n, lowlat300n, yhigh, ylow, highlng300n, lowlng300n);
highlat300n = highlat300n + 5;
lowlat300n = lowlat300n + 5;
highlng300n = highlng300n + 5;
lowlng300n = lowlng300n + 5;
} else if(tlng > lng && tlng < lng + 591999.2){
//point is within 400 meters on lng
//on the POSITIVE SIDE
layers.addcube(highlat400n, lowlat400n, yhigh, ylow, highlng400p, lowlng400p);
highlat400n = highlat400n + 5;
lowlat400n = lowlat400n + 5;
highlng400p = highlng400p + 5;
lowlng400p = lowlng400p + 5;
}else if(tlng < lng && tlng > lng - 591999.2){
//point is within 400 meters on lng
//on the NEGATIVE SIDE
layers.addcube(highlat400n, lowlat400n, yhigh, ylow, highlng400n, lowlng400n);
highlat400n = highlat400n + 5;
lowlat400n = lowlat400n + 5;
highlng400n = highlng400n + 5;
lowlng400n = lowlng400n + 5;
}else if(tlng > lng && tlng < lng + 739999){
//point is within 500 meters on lng
//on the POSITIVE SIDE
layers.addcube(highlat500n, lowlat500n, yhigh, ylow, highlng500p, lowlng500p);
highlat500n = highlat500n + 5;
lowlat500n = lowlat500n + 5;
highlng500p = highlng500p + 5;
lowlng500p = lowlng500p + 5;
}else if(tlng < lng && tlng > lng - 739999){
//point is within 500 meters on lng
//on the NEGATIVE SIDE
layers.addcube(highlat500n, lowlat500n, yhigh, ylow, highlng500n, lowlng500n);
highlat500n = highlat500n + 5;
lowlat500n = lowlat500n + 5;
highlng500n = highlng500n + 5;
lowlng500n = lowlng500n + 5;
}
}
如果有人能帮助我做我想做的事情更有效率，我会很感激！ 谢谢，
答案 0 :(得分：3)
一种方法是制作一个单元网格，并使用该点与原点的距离和单元格的大小来确定该点进入哪个单元格。计算单元格的公式是O（1） - 因此其效率不依赖于网格的大小。这是对某些代码的粗略攻击 - 未编译和未经测试，因此您可能必须对其进行修改才能使其正常工作：
public class SquareGrid
{
/** Lower, left corner of the grid */
private Point2D m_origin;
private double m_cellSize;
private List<List<List<Point2D>>> m_cells;
public SquareGrid(Point2D origin, int numberOfCellsPerSide, double cellSize)
{
m_origin = origin;
m_cellSize = cellSize;
m_cells = new ArrayList<List<List<Point2D>>>(numberOfCellsPerSide);
for (int i = 0; i < numberOfCellsPerSide; i++)
{
List<List<Point2D>> row = new ArrayList<List<Point2D>>(numberOfCellsPerSide);
m_cells.add(row);
for (int j = 0; j < numberOfCellsPerSide; j++)
{
row.add(new ArrayList<Point2D>>());
}
}
}
public void add(Point2D point)
{
int xCell = Math.floor((point.getX() - m_origin.getX()) / m_cellSize);
int yCell = Math.floor((point.getY() - m_origin.getY()) / m_cellSize);
m_cells.get(xCell).get(yCell).add(point);
}
}
如果您希望原点位于网格中间，则需要调整添加算法，并决定如何处理每边奇数个单元格。
专业人士可能会使用R-Tree或其他一些数据结构来构建有效的空间索引而不是网格。