Many Cartesian Trees

The key annoyance is that we are not given one Cartesian tree. We are given the union of $n$ Cartesian trees while the minimum value keeps getting pushed above everything else. That sounds cursed, but the process has one very useful symmetry.

After one full cycle, the only thing that mattered was the cyclic order of values. If a valid permutation is

$q_1,q_2,\ldots,q_n,$

then

$q'_i = q_i \bmod n + 1$

is also valid. This just rotates the value cycle, so the set $S$ does not change. Therefore we may choose a convenient rotation. We will force position $n$ to have value $n$, reconstruct the values downward, and output that permutation.

For every vertex $v$, define:

$Child(v)=\{u \mid (u,v)\in S\},$

so these are all vertices that are ever children of $v$ in some tree. Let

$L_v = \min Child(v), \qquad R_v = \max Child(v).$

Also define the possible parents of $v$:

$Par(v)=\{u \mid (v,u)\in S\},$

and keep their minimum and maximum indices too.

Now suppose we already know that position $v$ has the current largest remaining value. Because of the cyclic-rotation symmetry, we can pretend this value is literally $n$. We need to find the position of the predecessor value, i.e. the value just below it on the cycle.

The first important fact:

The predecessor of $v$ must be either $L_v$ or $R_v$.

Here is the reason. Let $x$ be the position whose value is just below $q_v$. If there are children of $v$ on both sides of $x$, say one child to the left and one child to the right, then when $x$ becomes high enough in the rotated priority order, one of those edges would be forced to attach through $x$ instead of directly to $v$. In Cartesian-tree terms: parents are chosen using nearest greater elements; a closer nearly-largest element sitting between a child and $v$ blocks $v$. So $x$ cannot sit strictly inside the span of children of $v$. It must be one of the extremes. Nice little ambush, honestly.

So there are only two candidates.

There are easy cases:

• If all children of $v$ are left of $v$, then the predecessor must be the leftmost child, $L_v$.
• If all children of $v$ are right of $v$, then the predecessor must be the rightmost child, $R_v$.

The only real case is

$L_v < v < R_v.$

Now one candidate is on the left and one is on the right. We need to tell which one is the true predecessor.

Use crossing edges. Say an edge $(a,b)$ crosses position $v$ if $a$ and $b$ are on different sides of $v$:

$(a-v)(b-v)<0.$

If $x$ is the true predecessor of $v$, then no parent edge of $x$ ever crosses $v$. Why? Since $v$ has the larger value, if a parent of $x$ were across $v$, then $v$ would be closer on that side and would block that parent. Cartesian trees do not let you ignore the giant value standing in the middle like a traffic cone.

For the other extreme child $y$, there will be a crossing edge. When the true predecessor $x$ becomes the root in some rotated order, $y$ attaches through $x$, and since $x$ and $y$ lie on opposite sides of $v$, that edge crosses $v$.

So in the split case:

• If the left candidate $L_v$ has no parent to the right of $v$, then $L_v$ is the predecessor.
• Otherwise, the right candidate $R_v$ is the predecessor.

Using precomputed extrema, this is $O(1)$ per value:

• $childMin[v], childMax[v]$ from incoming edges to $v$.
• $parentMin[v], parentMax[v]$ from outgoing edges of $v$.

Algorithm:

1. Read the matrix.
2. For every $s_{i,j}=1$, update:
   • children of $j$ with child $i$,
   • parents of $i$ with parent $j$.
3. Set current position $p=n$ and current value $val=n$.
4. Assign $ans[p]=val$.
5. To find $val-1$:
   • Let $mn=childMin[p]$, $mx=childMax[p]$.
   • If $mx<p$, move to $mn$.
   • Else if $mn>p$, move to $mx$.
   • Else the candidates are on both sides. If $parentMax[mn]\le p$, move to $mn$; otherwise move to $mx$.
6. Repeat until value $1$ is assigned.

The input is guaranteed valid, so the needed child set exists at every step before the last.

The preprocessing scans the whole matrix, so the complexity is

$O(n^2)$

per test case, with $O(n)$ extra memory besides the input string currently being read. Since the total $\sum n^2$ is bounded, this fits cleanly.