如何找到给定点的最近正方形

Question

I have a 2-D array of squares (square shape), Each square has 50 units length and x, y co-ordinates. The distance between the squares is 5 units. The x, y co-ordinates are the bottom left corner of the square. Now, given any point(x,y co-ordinates) how can i find the closest square to this point.

square **sq = new square*[10];
for(int i=0;i<10;++i){
    sq[i]=new square[10];
}

int m=0, n=0;
for(int i=0;i<10;++i){
    m=0;
    for(int j=0;j<10;++j){
        sq[i][j].setCoOrdinates(m,n);
        m+=55;
    }
    n+=55;
}

// Given a point (x, y) how can i find the index (i, j) of closest square to this point.

Answer 1

Let's rigorously define the definition of distance from a square to a point: "The minimum of all lengths from the point (x,y) to some point within the rectangle". This gives some clear rules for defining the distance to a rectangle. For any rectangle, if the point(x,y) is directly above, below, to the left of, or to the right of the sides of the rectangle, the minimum distance to the rectangle is the straight line drawn either vertically or horizontally through the point. For instance, if your point is (40, 90) and your rectangle's bottom-left is (0,0) (and it is a 50x50 square), you can draw a vertical line through your point and the distance is min(90-(0+50), 90-0)

If the point is not directly above, below, left of, or right of the sides of the square, then the closest distance is to the closest of the four corners. You can simply take the min of all the distances to the four corners, which can be found by using the distance formula.

Simply apply this logic to each of your squares in your array and you should be good to go. Time O(NM) where n is the number of rows of squares and M is the number of columns of squares, In other words. O(number of squares you have). Space O(1).

Answer 2

Lets start with the simpler problem: All rectangles are 55 units wide, ie there is no empty space between them...

The best container is the one that you do not need. There is a simple relation between i and j index of a square in the array and its m and n coordinates:

const int distance = 55;
int m(int i) { return i*distance; }
int n(int j) { return m(j); }

As the relation is linear, you can invert it:

int i(int m) { return m / distance; }
int j(int n) { return i(n); }

Using integer arithmetics is fine here, because we get the same result as if we had used floating points and then rounded down.

This is already the full solution for the simpler problem. Given coordinats m and n the closest square is at index i(m),j(n) .

Now lets introduce the gap:

cosnt int width = 50;
const int gap = 5;

Now we have to distinguish two possibilities: The given coordinats are inside a square or outside. When it is outside there are two candidates for the cloest square.

int i_with_gap(int m) {
     int i = m / distance;

     // point is inside the square?
     if (m % distance <= width) return i;

     // point is closer to square at index i?
     if (m % distance <= width+ gap/2.0) return i;

     // otherwise i+1 is closer
     return i+1;
}