Answer: $58+2 \sqrt{10}$
Solution: As shown in the figure, find the symmetric point $\mathrm{B}^{\prime}(-1,4)$ of point $\mathrm{B}(4,-1)$ with respect to $\mathrm{y}=\mathrm{x}$, the symmetric point $\mathrm{B}^{\prime \prime}(1,4)$ of point $\mathrm{B}^{\prime}$ with respect to the $\mathrm{y}$-axis, and the symmetric point $C'$ of point $C$ with respect to $\mathrm{y}=\mathrm{x}$.
Since the length of $\mathrm{AB}$ is fixed, by symmetry, $\mathrm{BC}+\mathrm{CD}+\mathrm{DA}=\mathrm{AD}+\mathrm{DC}^{\prime}+\mathrm{C}^{\prime} \mathrm{B}^{\prime}$, so the minimum value is the case when $\mathrm{AB}''$ is directly connected,
$$
\mathrm{AB} "=\sqrt{58}, \mathrm{AB}=2 \sqrt{10},
$$

Therefore, the shortest perimeter is $58+2 \sqrt{10}$.
55. In a Cartesian coordinate system, given points $A(-2,-3), B(4,-1), C(m, 0), D(n, n)$, then the minimum value of $\mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{AD}$ is $\qquad$ _.