"Shortest Queue" sends the item to the queue with the least currently in it. Since all of your queues have a maximum capacity of 1 this setting wouldn't actually do anything, since the queues with an item in them aren't available anyway.
To send the items to the closest queue (by distance) ordering the ranks of the connections should be enough to achieve that with the current logic.
If the positions of the queues might change often, you can use code like the following to send to the closest available regardless of ranking.
int bestindex = 0;
double closest_dist = 999999;
for(int index = 1; index <= current.outObjects.length; index++)
{
if(opavailable(current, index))
{
Vec3 dist_vec = current.outObjects[index].getLocation(0.5, 0.5, 0) - current.getLocation(0.5, 0.5, 0);
double dist = dist_vec.magnitude;
if(dist < closest_dist)
{
closest_dist = dist;
bestindex = index;
}
}
}
// Return closest available queue
return bestindex;
