19. Given the equation
$$
x^{2}+y^{2}=3|x-y|+3|x+y|
$$

The area enclosed by the figure is $m+n \pi(m, n \in \mathbf{Z})$. Then the value of $m+n$ is ( ).
(A) 18
(B) 27
(C) 36
(D) 45
(E) 54
19. E.

When $x+y \geqslant 0, x-y \geqslant 0$, i.e., $y \geqslant -x$ and $y \leqslant x$, we have $x^{2}+y^{2}=3 x-3 y+3 x+3 y=6 x$ $\Rightarrow (x-3)^{2}+y^{2}=3^{2}$;
When $x+y \geqslant 0, x-y > x$, we have $x^{2}+(y-3)^{2}=3^{2}$;
When $x+y > x$, we have $(x+3)^{2}+y^{2}=3^{2}$.
In summary, the curve represented by the equation is shown in Figure 6, and the area of the enclosed figure is
$$
6^{2}+4 \times \frac{9 \pi}{2}=36+18 \pi .
$$

Thus, $m=36, n=18$.
Therefore, $m+n=54$.