Probably just to make the problem statement a little bit more interesting. Obviously it was not necessary to work with the two arrays. You can replace them with one array called profit. Any many people have done this.

In Soviet Russia you do not solve problems, problems solve you

It should ignore some corner cases, but shouldn't be shitty like that. It should contains some large random test cases

I hate so weak samples in such tasks.

I would accept sample like "1\n1 2\n" if in task there is a permutation and there is a very easy slow (e.g. n!) solution, which I can write in one or two minutes to stress my solution.

But in non-trivial geometry task like this I should write very non-trivial slow solution and non-trivial generator to check my solution. It would take me for half of the contest, so, I just submitted this problem when I have passed samples. Fortunately, I rechecked my code and resubmitted it after few minutes, because I made a stupid bug, and samples didn't catch it.

On systems where you don't get any feedback about the correctness of your code during the contest (TopCoder, Facebook Hacker Cup, Google Code Jam, etc.), I think that the samples should be a bit larger so that you don't make a very trivial mistake that is caught on every case except the really small ones. However, for contests where you are given some feedback during the contest (ICPC, CodeForces, etc.), I think that weak sample data is perfectly fine because you are told that your program is wrong during the contest and then it is up to you to find out where you are wrong.

Weak examples for 250s (which usually contain corner cases for hacking) and strong examples for 500s.

No less than 8 samples was the rule I gave myself for my (one) SRM. Competitions like TCO final may be an exception, but I tend to differentiate between competitions and lottery. I'm not really interested in preparing the latter.

I'm sorry for the weak examples. We didn't intend to make a lot of solutions fail. What is intended (at least by me) is not to cover every corner case of the problems so that you have to think well the solution and then have chances to make challenges. I'm sorry if this bothered many of you.

I think that the samples always should be good, because solving of the problems is more important than challenging other solutions.

Yes

There is no pretesting. You can submit any compilable code.

First for every triangle with blue points as vertices compute number of blue and red points inside that triangle. Then fix lowest leftest point of the convex hull. Imagine that we build the convex hull. To add a new point we need to check the last 2 points that are already in the convex hull. So we sort all points by angle and calculate dp_i, j =  maximal number of blue points if the last 2 points in the convex hull are p_i and p_j. To calculate it we try all p_k with angle larger than p_i, check that cross product of points p_i - p_j and p_k - p_i is positive and check that triangle (o, p_i, p_k)(where o is lowest leftest point) does not contain any red points.

I discovered the bug for one of the members of my team. It is an issue with Greed, actually. It downloaded the inputs incorrectly. If you look at the BagAndCards.sample file, it is wrong--There are quotation marks around the string. I should point out that I also use Greed, but a newer version, and this did not happen to me during the contest, so you may want to consider upgrading. =)

Are you sure? FFT was my first thought and it looks like an intended solution (big TL).

The intended solution is something similar to FFT: instead of finding Fourier transform (calculating f(r^0), f(r^1), f(r^2).. where r is a unit root), we just find f()s at any m different values, like we just find f(0), f(1), .., f(m-1). Note that we don't need a 'fast' transform: we can just do it in O(m^2) for each of n vector. And we can precalculate how to restore answer from f(0), f(1), .., f(m-1).

Code: http://ideone.com/XFomMa

We intended to use another task that requires above idea, but then figured out that task have a simpler solution (but not like this one.. at least worth 600p). So we decide to change it to a new one just 2 days before the contest — and we just didn't notice this new task have such a simple solution.

I apologize for this issue and we will be more careful in the future.

The triangle ABD contains C (C is inside), so all 4 points are inside this convex polygon