Travel inside Square

Reflect the square and $Q$ several times as follows:

In order to meet the condition of one visit per side while traveling, the segment must pass through two horizontal square segments and two vertical square segments, and there are four possibilities as shown above: $PQ'$ , $PQ''$ , $PQ'''$ , and $PQ''''$ .

Using the fact that each square has a side length of $6$ and each $Q$ is $1$ unit vertically and $1$ unit horizontally away from the side of those squares, the coordinates of each reflected $Q$ are:

$Q'(-6 - 1, 2 \cdot 6 + 1) = Q'(-7, 13)$

$Q''(3 \cdot 6 - 1, 2 \cdot 6 + 1) = Q''(17, 13)$

$Q'''(-6 - 1, -2 \cdot 6 + 1) = Q'''(-7, -11)$

$Q''''(3 \cdot 6 - 1, -2 \cdot 6 + 1) = Q'''(17, -11)$

And each $PQ$ option has a length of:

$PQ' = \sqrt{(1 - -7)^2 + (3 - 13)^2} = \sqrt{164}$

$PQ'' = \sqrt{(1 - 17)^2 + (3 - 13)^2} = \sqrt{356}$

$PQ''' = \sqrt{(1 - -7)^2 + (3 - -11)^2} = \sqrt{260}$

$PQ'''' = \sqrt{(1 - 17)^2 + (3 - -11)^2} = \sqrt{452}$

The shortest possible path is $PQ' = \sqrt{164}$ , so $c = \boxed{164}$ .

The shortest path travelled from two points is one travelled by light. The shortest distance travelled from $P$ and $Q$ is one travelled by light where the four internal sides of the square serve as mirrors. Then the path from $P$ to the final image of $Q$ is a straight line. Let us consider the $x$ -component and $y$ -component of the absolute distance travelled (disregard the direction) by the beam of light.

$\begin{cases} |x_{PQ}| = \underbrace{|x_{PE}|}_1 + \underbrace{|x_{EF}| + |x_{FG}|}_6 + \underbrace{|x_{GH}| + |x_{HQ}|}_1 = 8 \\ |y_{PQ}| = \underbrace{|y_{PF}| + |y_{EF}|}_3 + \underbrace{|y_{FG}| + |y_{GH}|}_6 + \underbrace{|y_{HQ}|}_1= 10 \end{cases}$

Then the shortest path between $P$ and $Q$ $s = \sqrt{|x_{PQ}|^2 + |y_{PQ}|^2} = \sqrt{164}$ . Therefore $c = \boxed{164}$ .