Laser

Observation 1: the relative position never changes

The lasers always move by the same vector, so the offset between them is constant:

$$(\Delta x, \Delta y) = (x_2-x_1,\ y_2-y_1).$$

So the whole setup is just a rigid 2-point figure sliding around the board.

Also, any valid translated position is reachable: the set of translations that keep both lasers inside the board is a rectangle in translation-space, so you can move horizontally and vertically inside it without ever leaving the field. Nothing sneaky here.

Observation 2: which cells can one laser reach?

Take some cell $(x,y)$. The first laser can be at $(x,y)$ iff the second one, which must stay at

$$(x+\Delta x,\ y+\Delta y),$$

is also inside the board.

That means:

$$1 \le x \le n,\quad 1 \le x+\Delta x \le n,$$ $$1 \le y \le m,\quad 1 \le y+\Delta y \le m.$$

The number of valid $x$ is $n-|\Delta x|$, and the number of valid $y$ is $m-|\Delta y|$. So the first laser can melt exactly

$$(n-|\Delta x|)(m-|\Delta y|)$$

cells.

Same for the second laser. Symmetry doing its job, for once.

Let

$$a = |x_1-x_2|,\quad b = |y_1-y_2|.$$

Then each laser can reach

$$(n-a)(m-b)$$

cells.

Observation 3: those two sets may overlap

We need the number of cells reachable by at least one laser, so use inclusion-exclusion.

A cell lies in the overlap iff it can be occupied by both lasers. That means both shifted cells exist:

$$(x+a, y+b) \text{ is inside}, \quad (x-a, y-b) \text{ is inside}$$

up to signs, and only the absolute differences matter.

So horizontally, the cell must have room for both directions, which gives

$$\max(0, n-2a)$$

possible columns. Similarly vertically:

$$\max(0, m-2b).$$

Hence the overlap size is

$$\max(0, n-2a)\cdot \max(0, m-2b).$$

Final formula

The number of meltable cells is

$$2(n-a)(m-b) - \max(0, n-2a)\max(0, m-2b).$$

Therefore the number of cells that cannot be melted is

$$nm - \left(2(n-a)(m-b) - \max(0, n-2a)\max(0, m-2b)\right).$$

Why continuous movement doesn’t break this

The statement mentions moving continuously, with lasers active all the time. Sounds scary, but it changes nothing.

If a laser sweeps through some cell during a horizontal/vertical move, then the center of that cell is also crossed at some intermediate moment. Since all intermediate integer positions are valid stopping points as well, counting reachable cell centers is enough. No half-cell black magic needed.

Complexity

Each test case is just a few arithmetic operations:

$$O(1)$$

time and

$$O(1)$$

memory.

Given $t \le 10^4$, this is hilariously safe.