Of course it is - if you write the code to manage that.
If you are simply drawing stuff in a container from first to last, then it follows that any stuff later in the container will be on top of earlier stuff, so to find the one "on top" would be a simple matter of searching the container in reverse order and the first one to match the coordinates will be the top one.
More complicated scheemes can easily be conceived if you need something more complicated.
Edit: for the simple solution, look into rbegin(), rend().