This table is, indeed, calculated by assuming that a ship can produce constant acceleration away from its origin, instantaneously pivot \$180^\circ\$ at the midpoint of its journey, and constantly decelerate for the second half of its journey. The greatest veloticy thus attained (w.r.t. origin) in the table above is roughly 1 million m/s--approximately 1/3 of one percent of c--so we're comfortable sticking with classical calculations.
The formula that produces the above times is:
$$ t = 2 \times \sqrt{\dfrac d a} $$
where \$t\$ is measured in seconds, \$d\$ in meters, and \$a\$ in meters/seconds2. (G is rounded to 10 m/s2 for ease of calculation. Note that the table gives distances in km, so you've got to tack on three zeroes to end up in meters.)
Those who would like a refresher on their classical kinematics, read on:
Recall that the distance travelled when starting at rest and undertaking constant acceleration is given by
$$ d= \frac 1 2 at^2 $$
Solving for t gives us
$$ t=\sqrt{2 \times \frac d a} $$
In our situation we consider the time (t1/2) to accelerate to the midpoint of the journey (d1/2):
$$ t_{\frac 1 2}=\sqrt{\dfrac{2 \times d_{\frac 1 2}}{a}} = \sqrt{\dfrac d a} $$
Doubling this gives a total trip time of
$$ t=2 \times \sqrt{\frac d a} $$
