If $$$n$$$ is a perfect square number, we find that the square dyeing scheme is always optimal.The answer is $$$4\sqrt{n}$$$.

Otherwise,if $$$n-\lfloor \sqrt{n} \rfloor ^2 \leq \lfloor \sqrt{n} \rfloor$$$,the answer is $$$4\lfloor \sqrt{n} \rfloor+2$$$.Otherwise the answer is $$$4\lfloor \sqrt{n} \rfloor+4$$$.

If $$$n$$$ is odd,we can traverse all cells, which is a simple situation.

If $$$n$$$ is even,we can skip one of $$$(1,2) (1,4) (1,6)...(1,n),(2,1),(2,3),(2,5)...(2,n-1)$$$.If there is a duplicate number, skip it. Otherwise, skip the maximum number.

First,be careful when $$$y=0$$$ and $$$k=1$$$ :)

Consider the answer with a length greater than $$$n$$$, which is always $$$10...01...1$$$.The construction is not hard.

What about the answer with length $$$n$$$?

Consider setting numbers from the highest bit to the lowest bit and current bit is $$$i$$$.If $$$y_i=1$$$,we must set $$$x_i=1$$$.Otherwise($$$y_i=0$$$),we either do set $$$x_i=0$$$ and go to the lower bit,or do set $$$x_i=1$$$, $$$x_{i+1}x_{i+2}...x_n=0...01...1$$$ and break.

For $$$y_i=0$$$,when can we do set $$$x_i=0$$$ and go to the lower bit?The condition is $$$re1>pre1_{i+1}$$$ or $$$re1==pre1_{i+1}$$$ and $$$pre1_{i+1}==tot1_{i+1}$$$.$$$re1$$$ is the current remaining number of $$$1's$$$,$$$pre1_{i+1}$$$ is the length of the continuous $$$1$$$ prefix of $$$y_{i+1}y_{i+2}...y_n$$$,and $$$tot1_{i+1}$$$ is the total of $$$1$$$ of $$$y_{i+1}y_{i+2}...y_n$$$.

Note $$$maxdif(x,y)=max(k)(bit(x,k)!=bit(y,k))$$$.

It can be proved that $$$maxdif(a[i],a[i+1])$$$ is strictly decreasing.

Based on the above conclusion, we can make the optimal construction.

Note $$$x=(1...10???)_2$$$,construct $$$a=[(1...10???)_2,(1...110...0)_2,(1...111...0)_2,...,(1...111...1)_2]$$$.

Consider DP.

Note $$$dp_{i,j}$$$ as the maximum score we can get at the end of $$$(i,j)$$$.

Note $$$lmax_{i,j}=max(0,max_(1 \leq k \leq j)((a_{i,k}-2c)+...+(a_{i,j}-2c)))$$$ and $$$rmax$$$ similarly.

Now consider which cell we go from to $$$(i,j)$$$.There are four situations.

$$$1.$$$ We just start at $$$(i,j)$$$.The answer is $$$a_{i,j}+lmax_{i,j-1}+rmax_{i,j+1}$$$;

$$$2.(i-1,j) \rightarrow (i,j)$$$.The answer is $$$dp_{i-1,j}-c+a_{i,j}+lmax_{i,j-1}+rmax_{i,j+1}$$$;

$$$3.(i,j-1) \rightarrow (i,j)$$$.The answer is $$$left_{i-1,j}-c+a_{i,j}+rmax_{i,j+1}$$$;

$$$4.(i,j+1) \rightarrow (i,j)$$$.The answer is $$$right_{i+1,j}-c+a_{i,j}+lmax_{i,j-1}$$$.

$$$left_{i,j}=max(max(0,dp_{i-1,j}-c)+a_{i,j}+lmax_{i,j-1},left_{i,j-1}-c+a_{i,j})$$$,and $$$right_{i,j}$$$ similarly.

$$$dp_{i,j}$$$ is the maximum value of the answers for the four situations mentioned above.And the final answer is $$$max(dp_{i,j})$$$.