Divide and conquer algorithm-nearest point problem finding the closest pair of points

Problem Description:

Input: Point set Q output on space plane: two closest point pairs

Problem simplification: If you are looking for the nearest point pair in a straight line, you can use sorting and then find the nearest nearest point.

Divided treatment ideas:

Divide divides it into two parts q1,q2 T (n) = O (n)

Conquer find nearest point pair <p1,p2>,<q1,q2> T (n) = 2T (N/2)

Merge compare the distance between two points near the point of separation <p3,q3> and find the <p1,p2><q1,q2> distance t (n) = O (n)

Time complexity: T (N) =o (1) n=2

T (n) =2t (N/2) +o (n) =o (NLOGN) n≥3

Two-dimensional space problems:

The divide and conquer phases are similar to one-dimensional problems, with the X=m line divided into two sub-parts to find the nearest point to <p1,p2>,<q1,q2>

D=min{<p1,p2>,<q1,q2>}, the point of the merge stage must be within the [M-d,m+d] range.

There seems to be a lot of optimizations, but the worst case scenario is that the number of points within that range corresponds to a n^2 level of complexity.

More optimized scenario: For a point, there are up to 6 points on one side that are less than D. Specify the following

, the point P (x0,y0) in the left-hand area of the l:x=m, the point at which the distance is less than D is within a circle with a radius of D, at the center of P,

The portion of the circle to the right of the l:x=m must be in (m,y0-d) the lower-left vertex, (m+d,y0+d) within the rectangle of the upper-right vertex, which is within the red rectangular region of the figure.

When the rectangle is divided into 6 blocks, with a length of 2D/3 and a width of D/2, the maximum distance of points within each small rectangle is 5d/6 (diagonal), that is, each small rectangle has a maximum of one point.

Because if there are two points in the same rectangle, the minimum distance will not be d.

Therefore, there is a maximum of 6 points for each vertex in the [M-d,m+d] range to be judged.

Things to keep in mind:

1. To set up a good data structure for easy processing

2. To Preprocess, the point set is sorted by x, Y, respectively, and the two sort sorts are maintained together.

Time complexity: T (N) =o (1) n=2

T (n) =o (n) +2t (N/2) +o (n) =o (NLOGN) n≥3

Divide and conquer algorithm-nearest point problem finding the closest pair of points