Don't understand what is the profit of knowing the scroring distribution before contest. Almost all participants solve tasks starting from the most easy (= problem A).

if there is any number theory problem in this contest, there will be many hacks .....

I think the 5 minutes difference is just that famous DELAY!!

We found that in case of 5-10 minutes delay there are many participants who additionally register. It seems it is some kind of psychology: if you see that round will start on 19:30 (and even if you know about 19:25 as deadline for registration), your brain rounds 19:25 to 19:30 and probably you will be near your computer on 19:30, but not on 19:25.

PrinceOfPersia is in the same color, you know....

And your colour says your comments are trash, then?

Good luck to everyone

Did you create a new account for writing this comment? Why are you so afraid of using your original account if you are really thinking this way?

we've had enough math in the last few rounds...

I've come up with the following idea:

As you can see, we can associate a list of vertices with an index of an array (or map, if you will).

Step 1: create the vertex queues

Vertices on the diagonal (0, 0), (1, 1), (2, 2), (3, 3), (4, 4), ... are all map to 0. That means, we can store these particular vertices under the key 0 in some container. I store them like that:

map< vector< pair<int, int> > > vertex_queues;
vertex_queues[0].push_back( make_pair(1, 1) );
vertex_queues[0].push_back( make_pair(3, 3) );
vertex_queues[0].push_back( make_pair(2, 2) );
...

vertex_queues[0] now stores the vertices unordered. So, we need to sort them so that we can use them later.

Step 2: construct the answer by taking the vertices from the vertex queues in the right order

The last step is building the answer. We go through the array w[n] from left to right and take the first element from the vertex queue with the index w[i]. So, the i'th vertex must be the one we've just taken from the queue. One by one, we extract all the vertices from the vertex queues in the order dictated by the array w[n].

If there are no elements in the w[i]'th vertex queue, the anser is NO. If, lastly, we look though all the vertices in the array that we've constructed and find that the current vertex is less than the previous one, the answer is also NO.

Otherwise, we've build the perfectly valid array of vertices which is the solution to the problem :)

14288949 — with map
14289137 — with vector

First you should sort points by their y-x then for each w[i] from last to first set the point with biggest x and y!

now we should check that does this greedy algorithm make a aesthetically pleasant! sort points and then for each (x,y) we calculate the mp[ (x,y) ] the maximum index of every point like x' and y' that 0<=x'<=x and 0<=y'<=y in the answer! and we update it with mp[ (x-1,y) ] and mp[ (x,y-1) ] , if one of them was bigger than index of (x,y) print No!

now we know answer for each w[i] and it's aesthetically pleasant!

UB в C++ — вещь непредсказуемая. Могло и нормально отработать.

I had draft ready. Only posted it after contest finished.

I also wanted to have my first hacking attempt with such a silly mistake, that I also had made, but I had not found in time where the hacking button is, what a shame...

Nobody in my room was stupid enough not to use long long.

But reading the problem through the clutter is a part of the problem. If you would be said like "given array, find sum of absolute differences between i and i+1 for i from 0 to N-1" then it wouldn't be a contest.

Еще бы русский разбор и вообще отлично...

эти точки не являются вершинами прямоугольника положительной площади, у которого стороны параллельны осям.

Seems like it. dp[left/right][start][end] precompute two arrays to determine the first tree that will be available to cut if ith tree is cut and falls on left, and right respectively.

This straight line in your curve looks pretty cool though.

On the other hand, you got the 228th place, Bredor is very proud of you!

The computer component principle is a nightmare!!

The correct answer for that test is 227504, and yours is 0

Correct answer isn't 0. Its 227504. I failed at the same due to a stupid mistake. Should have tested more rather then racing against time.

Maybe he decided that Div.1 contests are too complicated for him and will participate only in Div.2?

I think it is because while loop in your find function.

numbertheorist17,thesilione

My AC algorithm fails it too

UPD: And also the winner's solution

Sorry, my mistake, w is negative not x,y

My submission is also wrong! :)

my submission works good even on this test (however I solved the question after contest !!)

You just made my day pretty sad T_T

Your answer does not actually specify the last significant digit of the answer

Thats strange. Even the basic test such as "1 1000000000" runs forever on my computer. Are there any kind of optimizations being done in codeforces compiler?

C++ Optimizer. Optimizer is much smarter than that programmer.

Your input:

3
6 3
7 0
6 2
-3 -7 -4

is not valid, is it? Take a closer look to the input section of problem C.

"Also, it is guaranteed that if some point (x, y) is present in the input, then all points (x', y'), such that 0 ≤ x' ≤ x and 0 ≤ y' ≤ y, are also present in the input."

I'll take a look.

Also my solution verify this :D (I'm 99.00% sure about my solution correctness)

This is unfortunate. I thought I took care of all the cases, but I guess I didn't. Sorry about that! Next time, I will have to be more careful.

MikeMirzayanov, GlebsHP take a look please

During the contest I got intimidated by the number of people who solved D and didn't give it much thought. Later I tried to solve it and found it easier than C either, it's a straight foward DP.

Actually I think the hardest task on C was understand the english translation, the problem itself wasn't hard.