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)? [Pre-computation]

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 $$$\texttt{for(auto x: C_i)}$$$.

Case 1 ($$$\texttt{Keep}$$$): If $$$x$$$ is not overlapping now, then we add $$$x$$$ to the answer, no matter if $$$x$$$ is from $$$A_i$$$ or not.

Case 2 ($$$\texttt{Remove}$$$): If $$$x$$$ is fully contained in the union of other rectangles, then we always remove $$$x$$$, no matter if $$$x$$$ is from $$$A_i$$$ or not.

$$$\texttt{Keep}$$$ and $$$\texttt{Remove}$$$ are relatively easier, no matter $$$x \in A_i$$$ or $$$x \in \{x_i \mid x \in B \}$$$.

Case 3 ($$$\texttt{Shrink}$$$): If $$$x$$$ is partially overlapping with other bricks.

Case 3.1: If $$$x \in A_i$$$: $$$\texttt{Shrink} x$$$. Move the $$$x.l$$$ to the right to avoid overlapping.

Case 3.2: If $$$x \in \{x_i \mid x \in B \}$$$: $$$\texttt{Shrink}$$$ all bricks that overlap with $$$x$$$ by moving their right endpoints (i.e., *.r), to the left.

It is guaranteed in Case 3.2 that every brick overlapping with $$$x$$$ is from $$$A_i$$$ because we have already performed pre-calculations in $$$B$$$ to make them non-overlapping. The complexity is not an issue, by amortized analysis, each $$$x \in A_i$$$ will do $$$\texttt{Remove}$$$ at most once and $$$\texttt{Remove}+\texttt{Shrink}$$$ at most twice (keep/move left endpoint+keep/move right endpoint).

Part 3: Implementation

First I copy a range-set implementation from nok0. Basically, it maintains a set of intervals using $$$\texttt{std::set}$$$. When inserting/erasing a range, it can merge/split/erase intervals automatically, using binary search (Yes, stop learning useless algorithms, go to learn binary search!). It supports many APIs, here are some useful APIs in our implementation:

(1) $$$\texttt{Insert(l, r)}$$$: Insert an interval $$$[l, r]$$$.

(2) $$$\texttt{covered_by(l, r)}$$$: Return an interval covering $$$[l, r]$$$, if not exists, returning $$$(-\infty, \infty)$$$.

(3) $$$\texttt{covered(l, r)}$$$: Return a bool variable indication whether interval $$$[l, r]$$$ is fully covered.

(4) $$$\texttt{right(x)}$$$: Return $$$\min \{y|y \geq x, y \text{ is not covered}\}$$$. Note that API could calculate the $$$\texttt{mex}$$$, so it is also useful for calculating the Sprague-Grundy (SG) number. SG number is irrelevant to our topic.

(5) $$$\texttt{left(x)}$$$ (added by me): Return $$$\max \{y|y \leq x, y \text{ is not covered}\}$$$.

There are also APIs that we don't use in this implementation, but they are also beneficial:

(6) $$$\texttt{Erase(l, r)}$$$: Erase $$$[l, r]$$$. That may lead to interval splitting, for example, if we erase $$$[2, 3]$$$ from $$$[1, 4]$$$, we get two intervals remaining: $$$[1, 1]$$$ and $$$[4, 4]$$$.

(7) $$$\texttt{size}$$$: Get the size of intervals.

(8) $$$\texttt{dump}$$$: Output all intervals for debugging.

So, we implement the pre-computation step (first paragraph in Part2) as follows:

sort B w.r.t l (left endpoint).
Initial a range set named rs.
for(auto& x: B){
    int rightl = rs.right(x.l), leftr = rs.left(x.r);
    if(right <= rightr){
        rs.insert(rightl, leftr);
        for(int i = 1; i <= 2; ++i)  {
             Ci = Ai; Ci.push_back({x.u, x.rightl, x.d, x.leftr, x.id}); //push into Ci
        }
    }else{
        //Case 2, fully contained, remove it!
        //We actually do nothing here
    }
}  

Then, how to calculate $$$C_i$$$ from each floor independently? We need another set that is sorted by $$$r$$$ in the ascending order, let's call it $$$\texttt{sortbyr}$$$, to maintain rectangles in $$$A_i \subseteq C_i$$$. If Case 3.2 happens, we need to find such $$$y \in \texttt{sortbyr}$$$ quickly by $$$\texttt{sortbyr.lower_bound(x.l)}$$$, i.e., find an interval $$$y \in A_i$$$ quickly, using binary search, such that the right endpoint of $$$y$$$ ($$$y.r$$$) is greater or equal than $$$x.l$$$. Note that when we find such an $$$y$$$, you need to iterate to the end of $$$sortbyr$$$, that's why my previous submissions WA at test15--I only handle one of them. Here is the sketch of the implementation:

for(int i = 1; i <= 2; ++i){
    Initialize sortbyr; //Remember to initialize because we handle each floor independently!
    sort(Ci.begin(), Ci.end());
    Initial a range set named rs.
    for(auto x: Ci){
         bool lcovered = rs.covered(l), rcovered = rs.covered(r);
         if(!lcovered && !rcovered){
             //Case 1, Keep
             if(u == d){
                //so
                sortbyr.insert({x.u, x.l, x.d, x.r, x.id});
             }
             ans[x.id].insert(node{j+1, x.l, j+1, x.r, x.id}); //No matter u == d or u != d, we only consider one floor here!
             rs.insert(x.l, x.r);
         }else if(lcovered && !recovered){
             if(u == d){
                 //Case 3.1
                 int rightl = rs.right(l);
                 ans[x.id].insert(x.u, x.rightl, x.d, x.r, x.id);
                 rs.insert(rightl, x.r);
                 sortbyr.insert(x.u, x.rightl, x.d, x.r, x.id});
             }else{
                 //Case 3.2
                 auto y = sortbyr.lower_bound({-1, -1, -1, x.l, -1});
                 while(y != sortbyr.end()){
                     //Here, we should deal with all overlaps! That's why my previous submissions WA on test15
                     auto [u2, l2, d2, r2, id2] = *y;
                     //Here l2 <= l because we sort Ci by l. So sortbyr only contains elements whose left endpoint <= l!
                     //By lower_bound, we have r2 >= x.l
                     ans[id2].erase({u2, l2, d2, r2, id2}); //We erase it from our answer
                     sortbyr.erase(y++);
                     ans[x.id].insert({j+1, x.l, j+1, x.r, x.id}); //We never shrink the splitted ones
                     rs.insert(x.l, x.r); //Although [l, r] overlaps with [l2, r2], range set will merge them automatically!
                     if(l2 <= l-1){ //Be careful! r2 may <= r, so y after shrinking may be empty!! 
                         ans[id2].insert({u2, l2, d2, x.l-1, id2});
                         sortbyr.insert({u2, l2, d2, x.l-1, id2});
                     }
                 }
             }//Case 3.2
         }//Case 3
    }//for
}