Lasers

The entire trick is that the coordinates look scary, but they are basically decoration.

We start at $(0,0)$ and need to reach $(x,y)$ inside the rectangle.

There are two kinds of barriers:

• horizontal lasers: from $(0,a_i)$ to $(x,a_i)$
• vertical lasers: from $(b_i,0)$ to $(b_i,y)$

A horizontal laser spans the full width of the rectangle. A vertical laser spans the full height. So you cannot sneak around the endpoint like some cartoon burglar. The endpoints touch the rectangle boundary, and your path must stay inside the rectangle.

Why every horizontal laser must be crossed

Take any horizontal laser at height $a_i$.

At the start, your $y$-coordinate is $0$. At the end, your $y$-coordinate is $y$.

The statement guarantees:

$0<a_i<y$

Your movement is a continuous curve, so your $y$-coordinate changes continuously from $0$ to $y$. By continuity, at some moment it must equal $a_i$.

That means your path touches the horizontal line $y=a_i$, which is exactly crossing that laser.

So every horizontal laser contributes at least $1$ crossing. There are $n$ of them, so that gives a lower bound of $n$ crossings.

Why every vertical laser must be crossed

Same exact argument, but with the $x$-coordinate.

A vertical laser is at $x=b_i$, and the statement guarantees:

$0<b_i<x$

Your $x$-coordinate starts at $0$ and ends at $x$. Since the path is continuous, it must equal $b_i$ at least once.

So every vertical laser contributes at least $1$ crossing. There are $m$ of them, so that gives another lower bound of $m$ crossings.

Together, any valid path needs at least:

$n+m$

crossings.

Can we actually achieve $n+m$?

Yes.

Use any path that moves monotonically from $(0,0)$ to $(x,y)$, for example a straight line segment. Along such a path:

• each horizontal laser is crossed exactly once,
• each vertical laser is crossed exactly once.

If the path passes through an intersection of a horizontal and vertical laser, the problem says that counts as two crossings. That is still fine: it is exactly one crossing for the horizontal laser and one crossing for the vertical laser.

So intersections do not reduce the answer. They also do not increase the minimum, because those two lasers had to be crossed anyway.

Therefore the minimum possible number of crossings is exactly:

$\boxed{n+m}$

The arrays $a$ and $b$ only need to be read from input. Their actual values do not affect the answer at all. Brutal 800-rated moment: read numbers, print the count.