Most efficient way of implementing this?

I've got a 1km square, I'm dividing it up into 100meters both on the positive and negative sides of it..

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

Basically I'm making a square just like the one above. I'm diving it up into 100 meter squares both on the positive side and negative side of the x axis. I'm then plotting each point in their respective co-ords , I'm planning on throwing an (x,y) value through a bunch of if/else statements (at least 50 , upto 100) but I'm concerned with how expensive this would be. Is there a more efficient way of doing what I want to do?

Here is an example of how I plan on doing it...

              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 POSITIV开发者_Python百科E
                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;
            }
            } 

If anyone could help me make what I want to do much more efficient I'd appreciate it! Thanks,

One way is to make a grid of cells, and use the distance of the point from the origin and the size of the cell to work out which cell the point goes into. The formula for calculating the cell is O(1) - so its efficiency doesn't depend on the size of the grid. This is a rough stab at some code - uncompiled and untested so you might have to modify it to get it to work:

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);
    }
}

If you want the origin in the middle of the grid you will need to adjust the add algorithm, and decide how you want to deal with odd numbers of cells per side.

The professionals would probably use an R-Tree or some other data structure that can be used to build an efficient spatial index instead of a grid.