Balanced circle packing
Here are examples of balanced circle packings whose touching graphs are disconnected:
Circles of same color are congruent and of equal weight. The drawing on the right shows how recursion can produce an arbitrarily large number of connected components. The very same number of circles and with the same arrangement pattern works in every dimension $d\ge2$; just replace each circle, including the container, with a concentric $d$-dimensional ball of the same radius.
Another example:
$D_1$ and $D_2$ touch the boundary of $D$ and each other, but $D_i$ (for $i>2$) touches neither the boundary of $D$ nor any of the other disks.
I would like to argue for the following claim: In $\mathbb{R}^1$, the disks in a tightest balanced packing form a tiling, i.e., there are no gaps between the $D_i$ disks.
First, observe that in a tightest balanced packing, either the left or right end of the container disk $D$ is in contact with a contained disk $D_i$. Let $L$ and $R$ denote the left and right ends of the disks in a tightest balanced packing. Let $c$ be the center of gravity of the weighted disks, and $c_D$ the center of the containing disk $D$. Because the packing is balanced, $c = c_D$. If $D$ extends both left of $L$ and right of $R$, then the diameter $d$ of $D$ can be reduced, keeping $c_D$ fixed. So $D$ must match either $L$ or $R$ (or both) in a tightest balanced packing. Without loss of generality, let $D$ match $L$.
Now, suppose in contrast to the claim, that there is a tightest balanced packing with at least one gap between the packed disks. See the figure above, (a). Fix the disks left of the gap, and move all the disks right of the gap in unison leftward slightly, reducing the gap width. This slides $c$ to $c'$ toward the left, and slides the right end to $R' < R$. In general the displacement $c-c'$ is less than $R-R'$. Now move $D$ leftward so that its center $c'_D$ matches $c'$; see (b) in the figure. Now $D$ contains the disks, matches the center of gravity, but does not touch a disk either on the left or to the right. It doesn't match $L$ because $D$ was moved leftward. It doesn't match $R'$ because the disks right of the gap moved leftward at least as much as the leftward movement of $c_D$ to $c'_D$. So the diameter $d$ of $D$ can be reduced. So the assumed tightest balanced packing was not tightest afterall.
This allows a simple exponential algorithm: for $n$ weighted disks, try all $n!$ arrangements of the disks, and select the one whose center of gravity $c$ is closest to the midpoint of their combined length. Then surround as tightly as possible with $D$'s center matching $c$. In general, $D$ will only touch on the left or the right, as in the example shown below.
$r_i=(.39,.06,.27,.29)$. $w_i=(.35,.18,.10,.37)$. $d=1.032$.