Problem: Rectangle Shrinking

Part 1: Notations

$$$x$$$: A brick. $$$x_i$$$ denotes the part of $$$x$$$ belonging to the $$$i$$$-th floor. For example, if $$$x.u=1,\,x.d=2$$$, then $$$x_1.u=x_1.d=1$$$, $$$x_2.u=x_2.d=2$$$, and $$$x_1.l=x_2.l=x.l$$$, $$$x_1.r=x_2.r=x.r$$$.

$$$A_i,\,i=1,2$$$: The set of bricks whose $$$u=d=i$$$.

$$$B$$$: The set of bricks whose $$$u=1,\,d=2$$$.

$$$\texttt{Remove}$$$: Remove a brick $$$x$$$.

$$$\texttt{Shrink}$$$: Shrink a brick $$$x$$$ to another non-empty brick.

$$$\texttt{Keep}$$$: Keep a brick $$$x$$$ unchanged.

Note that in our solution we may operate on a part of brick. For example, for a brick $$$x \in B$$$, we may $$$\texttt{Keep}\,x_1$$$ while $$$\texttt{Remove}\,x_2$$$.

Part 2: Idea

I think $$$\texttt{Shrink}$$$ing $$$x \in B$$$ is troublesome. If you shrink $$$x_1$$$, you have to shrink $$$x_2$$$ in the same way, i.e., $$$x_1$$$ has to be aligned with $$$x_2$$$. So is there a solution that only performs $$$\texttt{Remove}$$$ and $$$\texttt{Keep}$$$ on every $$$x \in B$$$? But that is impossible, because the bricks $$$x, y \in B$$$ may overlap themselves even with considering $$$A$$$. So what if $$$B$$$ are pairwise non-overlapping (because we could perform some pre-$$$\texttt{Shrink}$$$ and pre-$$$\texttt{Remove}$$$ to make $$$B$$$ pairwise non-overlapping)?

In that case, we can consider each floor independently. Let $$$C_i := A_i \cup \{x_i \mid x \in B \}$$$. For example, if $$$A_1 = \{[u=1, d=1, l=2, r=5], [u=1, d=1, l=7, r=10]\}$$$, $$$B=\{[u=1, d=2, l=2, r=2], [u=1, d=2, l=4, r=7]\}$$$, then $$$C_1 = \{[u=1, d=1, l=2, r=2], [u=1, d=1, l=2, r=5], [u=1, d=1, l=4, r=7], [u=1, d=1, l=7, r=10]\}$$$. We sort $$$C_i$$$ w.r.t $$$l$$$, and traverse $$$C_i$$$ like $$$\texttt{for(auto x: C_i)}$$$.