Then when moving from x=0 to x= 1, the distance of each retail store changes:
The distance between two retail stores with x less than 0 increased by 1, while the distance between four retail stores with x greater than 0 decreased by 1.
That is, the total distance is reduced by 2, then this movement is cost-effective.
So, where should I move to be the most cost-effective?
It should be moved to a place bigger than him and smaller than his retail store. As far as x is concerned, it is 3.
For example, if you move to 3, you have to move it again. If you move to 4,
There are two x's
So this kind of movement is not cost-effective, and it is not cost-effective to move from 3 to 2.
So x=3 is the most cost-effective.
Similarly, the most cost-effective y is 3 or 4.
Because (3,4) is already a retail store, only (3,3) is taken.