king only moves to the adjacent cell. it was in the statement.
also in chess you cant kill a piece with a king if it is defended by another same color piece.

what a pity!just go go!don't fuck,just work more hard and hack others!

Maybe you did not understand what "hacking" means. It is part of the contest. "Your solution was hacked" means that your solution was wrong and that someone found a test on which your solution fails and inputed it. This is part of the contest. So now, why are you angry that you got hacked, your solution would have failed the System tests eventually...

Одна из идей такая: так как все гамильтоновы пути проходят по каждой вершине один раз, давайте взвесим сначала не ребро, а вершину, а вес ребра будет суммой весов инцидентных ему вершин. Тогда если все ребра получатся разными по весу, то это и будет правильным ответом. Так как вес любого гамильтонового пути будет равен удвоенной сумме весов всех вершин. Остается только сгенерить веса вершин. Это можно попробовать сделать жадно. Дать вес 1 первой вершине. А дальше всем последующим вершинам давать такой минимальный вес, чтобы не получилось два ребра одного веса. Для 20 вершин, вес 20-ой вершины получится чуть больше 400. Поэтому веса ребер будут не больше, чем 1000.

did your solution failed on pretests?

Yes, it is strange(although probably fault is mine and not of the systest as always). Maybe some of you knows what the test 7( not pretest) is? I failed on that one.

If someone has test 7 for problemC(yes, 7 was my unlucky number today) It would be great.

Thanks in advance, and thanks to the setters for the nice problems!!!!

И ещё, пожалуйста, 21-й для B.

У меня там WA7, т.к. я зачем-то поставил на доску чёрного короля, и при генерации полей под боем сквозь него ничего пройти не могло (т.е. он мог просто сделать шаг назад от ладьи). Наверное, тест именно на это.

Да, тест 7 такой: a1 a2 c4 c2

Не знаю, что там в 7ом тесте, но я всех челленджил тестом "2 1 1 1" (на нём жадность не работает). Учитывая, что 7ой упал много у кого, может, это именно он?

А можно узнать, что за метод?
Я сделал просто рекурсией с запоминанием, и все прошло.

hahahaha. He's just to good to stand in that chart......

Выше уже был, читайте внимательнее!

N = 12, по-видимому ;-)

My solution got failed, but anyway I'll describe it( maybe someone can challenge it).
I think greedy works here.

-If you have an even number > 1 between two odd numbers -> -1
-If you have an odd number > 1 between two even numbers -> -1

If you don't then you must either:
1)do a division, or
2) a sum and a division
for any pair of numbers that need it.

repeat this until you end with 1 1 1 1

Here is the greedy I used (passed, but I proved it by bruteforce):
While you have a number different from 1:
1) Find the maximal number.
2) If it is even and has an even neighbour, divide. Otherwise, if it is odd and has an odd neighbout, increase them both, otherwise increase and any of it's neighbours.

Why does it work: Each time before you divide, the total sum increases by 4 (after you make 2 additions, you're guaranteed to divide). So, with each division the total sum decreases (compared to the sum after the previous division) unless the biggest number is 4 and is surrounded by 2 odd ones. Those are not many cases, I just wrone the program and tested all of  them - received answers on all. If anyone can give a better proof, you're welcome.

Bad System Tests? Maybe hacking tests aren't added to the System tests and that, along with a bug(Codeforces Beta - Contest alfa) gave the result?

MBabin didn't lock his solution, so after tourist had hacked it and got the score, MBabin just found the bug, fixed it and resent his solution.

You can resubmit solution after hack (if you haven't locked it already). So MBabin noticed that his solution is hacked, fixed bug and submitted new version.

Пожалуйста: h7 g7 h6 h8

#6:
c1 c2 d1 f1
#12:
b3 a8 c2 a3

В очередь! ))

a6 a8 c2 a1

The idea is to assign some numbers a_i to vertices and the lengths a_i + a_j to edges. Then the condition about hamiltonian cycles will be fulfilled. To ensure the condition about different lengths, just choose the numbers a_i one by one so that it was always true that a_i + a_j ≠ a_i' + a_j'. In other words, when you choose a new number, it mustn't be equal to a_i + a_j - a_k and a_i - a_j - a_k, for any already chosen a_i, a_j and a_k. It turns out that if n ≤ 20, this process never needs numbers more than 500, so all edges have lengths less than 1000.

We've put a link to the problemset analysis.