I also solved it in the aftermath. The factor which makes the solution unique is column 1.
You have to connect the upper part with the bottom part. You can't do it on the right hand side because of the pait (11). If you do it in column 1 if you blacken R7 and R8 you lose the opportunity to blacken R10 and R11.
If you blacken R7, R10 and R11 you blacken 3 cells insted of 2, and you can still connect top and bottom through R8C7 and R6C10, and this also gives a unique solution for the (11) pair in R3C11

For Mad Max, I did not get as much in the top as rob has in that image, but I had the rest. When I was stuck there I decided I would try to shade in all the unclued squares, save for the top left where at least one had to be left white, and see if that worked. Wasn't 100% sure though when I was done, since it could have been possible to squeeze out more by leaving more squares at the intersections of clues unshaded. I eventually convinced myself that there was no way a puzzle in a timed contest could be that cruel to have the valid and fairly well-constrained solution I had found and not have it be right, but it was a doubt I had even as I went to claim time bonus...

By the way rob, why didn't you shade R10C6 and R10C11? =P

where can i see the cumulative scores of all 3 tapas held till now.

Thanks rob, I hadn't noticed the Tapa bit was that much constrained. I was expecting more of maximise coloured unclued cells vs minimise double clued cells. But it seems the optimise bit was much smaller in this puzzle.



http://logicmastersindia.com/TVC/2012.asp

Yes, I really should have that pattern down by now.