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Обсуждение задачи 1367. Конфиденциально!Deepesson What is the answer for this testcase? [2] // Задача 1367. Конфиденциально! 9 янв 2021 01:49 #+-+-+-+-+-+ +-+-+-+-+-+# Edited by author 09.01.2021 01:54 Deepesson Re: What is the answer for this testcase? // Задача 1367. Конфиденциально! 9 янв 2021 02:37 Ok, stupid mistake again. Basically, the last line will be always \n So if you did something like this to read the input: std::string s; while(!std::cin.tie(0)){ std::cin >> s; } The Program will read an empty string. So you need to read as: while(!std::cin.tie(0)){ std::cin >> s; if(!s.size())break; } I got WA #6, So I thought my diagonal code was wrong! Lol. Edited by author 09.01.2021 02:47 Edited by author 15.01.2021 03:31 Edited by author 15.01.2021 03:32 Edited by author 15.01.2021 03:32 Deepesson Re: What is the answer for this testcase? // Задача 1367. Конфиденциально! 9 янв 2021 02:38 Btw the correct answer is: 11 11 Denis Koshman Precautions // Задача 1367. Конфиденциально! 24 авг 2008 05:35 Some implementation bugs I had 1. Some secret may become reachable several times from different sides when filling some area. Make sure that list of reachable secrtets does not overflow due to that. 2. If you test touching the secret along with testing walls, make sure you don't write something like that if(!has_wall(...) || !has_wall(...)) where has_wall() returns true/false along with adding newly discovered secret to the list. The danger here is that if 1st operand of || is 'true', the 2nd one is not even evaluated according to C++ standard. So it might happen that some secret is not added to the list because 'has_wall' is not called due to free passage on the neighboring wall. But it's safe to replace boolean || with binary |. The latter is always evaluated for both operands even if the 1st one returns -1 (all ones). Michael Rybak (accepted@ukr.net) I think the problem statement is slightly incorrect (+) [1] // Задача 1367. Конфиденциально! 21 апр 2006 19:25 It says: The Secret is a regular junction (+) with one of the objects of special interest contained inside. This object of interest is accessible from any of the four sides of the junction. As far as I understand, "the four sides of the junction" means left, right, up and down. In this case, the answer for the following test: ..+.. ..+.. ++#++ ..+.. ..+.# should be 10 01 because there's no way we can reach the middle Secret. Meanwhile, a program that outputs 11 11 gets AC, while the one that outputs what I think is correct gets WA at 6th test. The clarification that may help people having this problem is: one can access the Secret object not only from those 4 sides, but also diagonally. Denis Koshman Re: I think the problem statement is slightly incorrect (+) // Задача 1367. Конфиденциально! 24 авг 2008 04:56 Or those 4 sides are diagonal sides svr Test 2 Incorrect! [1] // Задача 1367. Конфиденциально! 4 дек 2006 00:48 My program solves all testes on forum but has WA2. After some experiments I established that test 2 has form : xx+xx x#|#x where symbols x are not important It is obvious that secrets # mutually unreacheable But experiments say that test has it's answer 11 11 svr Re: Test 2 Incorrect! // Задача 1367. Конфиденциально! 17 дек 2007 15:55 Reading ЧАВО helped. Was mistake of input. But problem has appeared rather difficult. Approach with Dfs from each # leads to TLE. It is necessary take to acount topology. Field consists of some lakes closed by + and #. We should start Dfs from inside of these lakes and #-s found this time on boundary are mutually achievable. AC!. P.S. It helpfull to make relation of approachibility having transitiveness property to be equivalence. It can be made by dividing one big cell in 4 smaller ones and moving along them. Edited by author 20.12.2007 07:56 Anatoliy 'Tolyan_NO' Tolstobroff What answer for... [8] // Задача 1367. Конфиденциально! 21 авг 2005 23:32 #+# +#+ #+# and #|# -#- #|# and #-# +#- #+# Vedernikoff Sergey Re: What answer for... [7] // Задача 1367. Конфиденциально! 22 авг 2005 02:41 My AC program outputs: 10100 01100 11111 00110 00101 and 10100 01100 11111 00110 00101 and 11100 11100 11111 00110 00101 Anatoliy 'Tolyan_NO' Tolstobroff Re: What answer for... [6] // Задача 1367. Конфиденциально! 23 авг 2005 02:48 I check this test but my program get Wa21. WHO TAKE wa21 Please help me or give some good tests. Anatoliy 'Tolyan_NO' Tolstobroff Re: What answer for... [5] // Задача 1367. Конфиденциально! 24 авг 2005 02:33 Asked me answer for this ### ### ### please. Vedernikoff Sergey Re: What answer for... [4] // Задача 1367. Конфиденциально! 25 авг 2005 19:34 My AC program outputs: 110110000 111111000 011011000 110110110 111111111 011011011 000110110 000111111 000011011 Anatoliy 'Tolyan_NO' Tolstobroff Re: What answer for... [3] // Задача 1367. Конфиденциально! 25 авг 2005 22:45 thanks. Now i got Wa22 What answer for #-# and fo # | # Tolstobrov_Anatoliy[Ivanovo SPU] Not need help more. [2] // Задача 1367. Конфиденциально! 25 авг 2005 23:23 I got AC I am very Happy. I find error. this test ##. #.. ..# dimozzz Re: Not need help more. // Задача 1367. Конфиденциально! 18 мар 2007 22:54 Answer is : 1111 1111 1111 1111 ??? 107th And what right for your test?? // Задача 1367. Конфиденциально! 25 авг 2007 14:01 I did it! Who has WA 22, remember that "For the sake of simplicity, you can assume that the photographed area is surrounded with the barrier of "+"-es." Edited by author 25.08.2007 19:52 ahyangyi_newid Very easy problem. Why so few people got ac? [2] // Задача 1367. Конфиденциально! 8 май 2006 09:11 lj_860603 Re: Very easy problem. Why so few people got ac? [1] // Задача 1367. Конфиденциально! 8 май 2006 16:57 I am so sorry that I don't know what is ac?This is my first time to come here.Please tell me.Thank you! DHSP Group Re: Very easy problem. Why so few people got ac? // Задача 1367. Конфиденциально! 8 май 2006 17:04 AC == accepted TLE == time limit exceeded MLE == memory limit exceeded CE == compilation error WA == wrong answer That's what I knew ^^. N.M.Hieu Edited by author 08.05.2006 17:07 Gigzzz(gigz@inbox.ru) Nice problem [7] // Задача 1367. Конфиденциально! 4 май 2005 04:56 Realy fun problem:) Memory/Time limit ballance interesting, somebody can solve it without using STL(i mean not using something like deque)? asd Re: Nice problem // Задача 1367. Конфиденциально! 5 май 2005 13:10 <% print('sdf'); %> Edited by author 05.05.2005 13:15 Kit Re: Nice problem [4] // Задача 1367. Конфиденциально! 10 май 2005 01:37 What you mean by STL? Are not you solve it by filling in a square table? Memory limit (in my opinion) is sufficiently loose. Gigzzz(gigz@inbox.ru) Re: Nice problem [3] // Задача 1367. Конфиденциально! 10 май 2005 13:38 STL - standart template library I fill square table, but i have problem with memory to store queue. Kit Re: Nice problem [1] // Задача 1367. Конфиденциально! 13 май 2005 14:09 If you want, you may write me on klukva2@mail.ru and I will tell you about my solution. It also will be interesting to see your one. Tolstobrov_Anatoliy[Ivanovo SPU] i got Wrong Answer #21 // Задача 1367. Конфиденциально! 6 авг 2005 00:28 Help me please!! Vedernikoff Sergey Re: Nice problem // Задача 1367. Конфиденциально! 13 авг 2005 00:55 I've solved it in Pascal (and, therefore, without STL). I don't see here any TLE/MLE balance - I simply used restricted queue. Only problem in this task - hard input format (two WA#1). Burunduk1 Re: Nice problem // Задача 1367. Конфиденциально! 10 май 2005 04:21 I think many people always solve problems without STL.
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