There was similar problem on Yandex finals 2011 (link). Solution on Russian is explained here.

B — Let's say that 2 points are connected if they can be paired. it's possible to pair 2*N points iff each component of such graph has even size (I don't know proof). To make number of edges linear connect each point only with 2 closest points in each direction (not 1 point because this point can be deleted). Now we need to check if sizes of all components are even if we delete a vertex. This can found with similar approach to finding articulation vertices.

Code

J — I have some shitty dp, but some team solved it very fast so I believe that there exists neater solution.

My code. It generates correct board for every prime k with size k² × k² and k³ zeros.

The solution is based on Sprague–Grundy theory. Let's solve the problem where we have only one bean:

We are going to calculate the grundy function for each i. According to the theory:

G_i = mex(G_i - Ci, G_{i - Ci + 1}, ..., G_i - 1), where mex is the minimum non-negative integer, that doesn't appear in the set (e.x. mex(0, 1, 4) = 2 ).

To find it on O(logn) I used a segment-tree, where I kept for each number in range [0, 10⁵] it's last appearance in G. Having this information, we run a simple recursion to find the minimum number, which last time appeared earlier, than i - C_i. It is our G_i, now we update the tree.

After calculating G, let's calculate the answer. According to the theory, G_game is the xor-sum of all G_i, such that A_i is odd. The answer is "Second" if G_game is equal to zero, otherwise "First".

code

Идея правильная, но есть тонкости. А именно, когда появляется новое покрашенное ребро, нужно не забыть из обоих его концов попробовать по нему пойти из всех доступных цветов, а не только тем цветом, которым мы сейчас по нему пришли.

http://e-maxx.ru/algo/kuhn_matching

E
You need 3 facts (I'll list them without proof). First, our trajectory has period g = gcd(h, w). Second, if the trajectory period has u up moves (along h side) and r right moves, then gcd(u, h) = gcd(r, w) = 1. Third, any trajectory satisfying these constraints is good, so the answer is , where gcd(u, h) = gcd(g - u, w) = 1.

F
Note that every empty cell that's neither start nor finish has exactly one vertical and one horizontal connection. This means that for every line (row or column) we can go from one side and greedily pair empty cells with connections (this implies the number of empty cells is even). This works for all lines that contain no end points and gives a hint where finish can be located: if there's 0 lines with odd number of empty cells, it is in the same line as start, if there are 2 — start must lie in one of them and finish in the other, and if it's more than 2 — the answer is NO. We have O(N) candidates for finish, we just check all of them and construct the path in O(N²) for each. I heard this solution can even be improved to O(N²) if we build DSU over the paths once for every line where we check the finish position.