Cottage Village

Observation 1: kill the geometry, reduce to intervals

Each house is a square centered on the $x$-axis, with sides parallel to the axes. So house $(x_i,a_i)$ occupies

$$\left[x_i-\frac{a_i}{2},\ x_i+\frac{a_i}{2}\right]$$

on the $x$-axis.

Because every square is centered on the $x$-axis, their vertical projections always overlap around $y=0$. That means two houses overlap in 2D iff their horizontal intervals overlap with positive length.

So yes, the whole problem is just a 1D gap-counting problem. The geometry was mostly there to look scary.

Observation 2: only neighboring houses matter

Sort all houses by their center $x_i$.

Any valid position for the new house must be one of these:

1. to the left of the leftmost house,
2. to the right of the rightmost house,
3. between some adjacent pair of houses.

There is no point checking non-adjacent pairs: if something lies between house $i$ and house $j$ with another house in between, that middle house is the real obstacle.

Observation 3: what positions inside a gap can work?

Let the new house have side $t$. Its projection on the $x$-axis has length $t$.

Consider two adjacent houses $i$ and $i+1$. Their free gap is

$$g = \left(x_{i+1}-\frac{a_{i+1}}{2}\right)-\left(x_i+\frac{a_i}{2}\right).$$

The new house must:

• fit inside this gap without overlap,
• and touch at least one existing house.

Inside a gap, there are only two possible touching placements:

• touching the left house: $$c_L = x_i + \frac{a_i+t}{2},$$
• touching the right house: $$c_R = x_{i+1} - \frac{a_{i+1}+t}{2}.$$

Now compare $g$ with $t$:

• If $g<t$, the new house cannot fit. Add $0$.
• If $g=t$, both positions coincide. Add $1$.
• If $g>t$, they are distinct. Add $2$.

That’s the entire combinatorics. No DP, no segment tree, no black magic.

Observation 4: don’t use fractions if you don’t have to

Everything involves halves, which is annoying but unnecessary.

Multiply the gap formula by $2$:

$$2g = 2(x_{i+1}-x_i)-a_i-a_{i+1}.$$

So for each adjacent pair we compute

$$d = 2(x_{i+1}-x_i)-a_i-a_{i+1}$$

and compare it with $2t$:

• $d<2t \Rightarrow +0$
• $d=2t \Rightarrow +1$
• $d>2t \Rightarrow +2$

Observation 5: the two outside positions always exist

There is always exactly one valid position touching the leftmost house from the left, and exactly one valid position touching the rightmost house from the right.

So we start with

$$\text{ans}=2.$$

Then we process all $n-1$ adjacent gaps.

Correctness sketch

After sorting, every valid placement lies either outside all houses or in a gap between adjacent houses. The two outside placements are always valid and unique.

For a fixed adjacent gap:

• any valid placement must touch at least one of the two bordering houses, so only the two touching positions matter;
• if the gap is smaller than $t$, neither works;
• if the gap is exactly $t$, both touching positions coincide;
• if the gap is larger than $t$, both are distinct and valid.

Summing these contributions over all gaps counts every valid center exactly once.

Complexity

Sorting dominates:

$$O(n\log n)$$

Time complexity is $O(n\log n)$, and memory usage is

$$O(n).$$

For a Codeforces $1200$, this is pleasantly straightforward once you stop letting the squares bully you.