More details?

It is guaranteed that aj = 1 for some j (1 ≤ j ≤ n). The answer <= 4000 => upper bound is 400000

Actually you can use upper bound = 4000 as you can move both up and down, so you can calculate answer in such a way that all middle values do not exceed max(k, 2·max(a_i)).

We can find dp[x] with binary search

We assumed that since derivative is linear then optimal solution can be found incrementally (optimal point for dp[x] shouldn't be less than optimal point for dp[x + 1]) or with ternary search.

We can calculate dp[x] linearly.

Lets define z as optimal size of the squares that we need to cut from corners of our square of size x to get optimal result dp[x].

We can notice that z can only increase if we increase x. So when we calculate dp[x] we can keep this value z that is currently provides best score for x - 1, we can push z to the right while it improves our answer. So z will be pushed right less than 10⁶ times.

J code is here. This runs in 215 ms.

We have quite complicated solution, I wander if there exists a simpler one.

Find all blocks of the graph. If there exists block with vertices of all three colors then answer is YES. You can always find necessary cycle. For example, it's possible to find an edge with two endpoints of distinct colors, find a vertex with the third color and find a cycle containing that edge and that vertex (as I remember there is a theorem that a graph is 2-vertex-connected if and only if for each edge and vertex of the graph there exists simple cycle going through the edge and the vertex).

Cycle can be found with maxflow if the vertex is source and endpoints of the edge are sink. Complexity will be O(maxflow·(M + N)) = O(2(M + N)) = O(M + N).

If no such block exists answer is NO.

If a tour exists, it will be contained in a single block (biconnected component). Hence we process each block separately. If every block contains at most two types of breweries, the answer is NO.

Otherwise take vertices a, b, c of different types of breweries and compute two internally vertex disjoint paths between every pair of them. If they contain the third type of brewery, we join the two paths to a cycle to get a tour.

Otherwise (this part actually never executed on the judge) just remember one path between every pair. Our goal is to make these 3 paths internally vertex-disjoint. If we take two paths, then they share one vertex X out of and the other two endpoints Y, Z differ. Let U be the vertex furthest from X that is contained in both of the paths. Note that U has the same brewery type as X. Taking the subpaths from Y to U and form Z to U makes them vertex disjoint without loosing any brewery type. Repeat this for all pairs.

You can clip not tetrahedra, but facets of the iceberg (it is much easier), then you have to compute the volume of a convex polyhedron defined by its facets, which is straightforward.

That's my approach as well (integrate piecewise quadratic function). To avoid the issues you mention, I just used segments [i;i+1] (since z is between -100 and 100) and integrated using three points i+0.25, i+0.5, i+0.75, to avoid hitting any vertices. And to find the cross-section I just intersected segments collecting all pairs of points with the plane and found convex hull. That seems to make the code quite sane, modulo convex hull with floating-point inputs.

Bruteforce precalc works.

Yet another representation: if we think of target string as concatenation of strings A = "01", then operations mean that we can push each A to left for 1 position (first push is creation). we can see that every state of process can be described as point (l₁, l₂, l₃, ..., l_k), k — number of A's, and l_i — number of pushes of A_i to left and l_i > l_i + 1 must be satisfied for every i. Hence, the whole process is a path from (0, 0, ...0) to (n - 1, n - 3, ..., 2 - n%2). So You are to calculate the number of specific paths. Actually, the reason that first answers are Catalan numbers (1, 1, 2, 5) is that it counts the number of paths not exceeding diagonal.