Contest starts at 3am here. Is there anybody from east Australia, east Russia, or New Zealand?

I had the exactly same problem. I spent my last 1.5 hours on B even though I had the idea for D, but it initially thought the code would be easier. B was really annoying... C and D though

You can solve it in NlogN with a segtree storing the following information:
Answer , Answer without last element , Answer without first element , Answer without first as well as last element.

This is my solution: http://ideone.com/Wdrr2g
Though untested, since I could not debug it on example cases(1 char mistake :( ) — the logic is exactly same to most of accepted solutions — you keep 4 things in segtree nodes, matchings in [l+1,r] in [l+1,r-1], [l,r-1] and [l,r] in [l,r] . just see the merge function its the whole solution..

The key observation is that there're only two ways to match two adjacent pairs: the edges are either parallel or they cross each other (it looks either like two parallel lines or as an X letter).

Let's store f(is left open, is open right) in each node of the segment tree (it denotes the number of ways to match everything inside the segment so that the state of the left and the right border corresponds to the boolean parameters (it's true if and only if the edge goes across)). We can merge two nodes by iterating over all possible parameters' values for the left and for the right node (in fact, there're 8 valid combinations as the middle ones must match).

The time complexity of this solution is O(N log N).

Aren't all angles were 45 degrees if you wanted to achieve minimal cross section? Meaning from every pole it was going down at 45 degrees both left and right unless some other pole prevented it from happening?

I just multiplied height and position by 2 and did everything even in integer numbers.

Are there cases where something else then 45 degrees gives optimal solution? if this is the case then this must be my error in understanding the problem statement.

It's more like data structures, and not really a geometry :)

Don't know about zholnin, but I simply split each 1x1 cell in 4 parts by 2 diagonals, and after that putting a pole is equal to labeling some of these triangles as used.

It would be very helpful if you can give me current area + sum of current areas after each pole. I have this:

49.000000 49.000000
2549.000000 2500.000000
6050.000000 3501.000000
12386.000000 6336.000000
19046.000000 6660.000000
25742.000000 6696.000000
32678.000000 6936.000000
39694.000000 7016.000000
47058.000000 7364.000000

Case #4: 47058.000000

And this is the answer at the bottom.

I had the same idea but I only used (n,m) as (max, min) ... Now I think about it that was very stupid.

I could not debug it on example cases — had 10 minutes left when i started to debug it the hard way. you keep matchings in [l+1,r], [l,r-1], [l+1,r-1], and [l,r]. Base cases of all others are 1 except [l+1,r-1] — its 0. Code — http://ideone.com/Wdrr2g
Edit — It matches the input/output with accepted solutions :(

Finding the area of building and clouds were standard for me but I couldn't find a way to compute the area of water, at least during the contest — this is why I liked it. Maybe I missed something and it's standard indeed.

For me C was yet another segment tree problem with complicated merging. It's so known that I used a similar problem on a CF round before (573D - Bear and Cavalry), but that one required some thinking first and otherwise I likely wouldn't use it at all.

Yeah looks like one of us wasted time :)

At least I saved some time by not competing in the last hour.

The red one is 2014's and the other is 2015's. The 2016's one had the same design but the color was pale grey. The worst thing is the writings on the back they write lots of stuff with large fonts.

This is on codejam tshirt :P FBHC writes something like "Code fast and win things" with a link to their facebook page on the tshirt -_-

Yep, thanks for confirmation.