這是最有效的方法嗎？

Question

我有一公里多，我將它分成兩百米在它的正面和反面..

  __________y__________ 
     |   |   | 
     |   |   | 
     |   |   | 
     |---------|---------| x 
     |   |   | 
     |   |   | 
     |_________|_________| 

基本上我正在做一個正方形就像上面的一個。我在x軸的正方向和負方向上將它分成100米的正方形。然後我繪製每個點在他們各自的合作伙伴中，我打算通過一堆if/else語句（至少50，最多100）拋出一個（x，y）值，但我關心的是如何這將是昂貴的。有沒有更有效的方式來做我想做的事情？

這裏是我計劃如何做一個例子...

   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; 
      } 
      } 

如果有人可以幫助我做什麼，我想要做的更有效的我會感激！ 謝謝，

Answer 1

一種方法是做一個網格的單元格，並使用距離原點和單元格大小的點的距離來確定點進入哪個單元格。計算單元格的公式是O（1） - 因此它的效率不取決於網格的大小。這是一些代碼粗糙刺 - 未編譯的和未經考驗的，所以你可能需要修改它來得到它的工作：

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); 
    } 
} 

如果你想在網格中間的起源，你將需要調整加算法，並決定如何處理每邊奇數個單元格。

專業人士可能會使用R-Tree或其他可用於構建高效空間索引而不是網格的數據結構。