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| Show all threads Hide all threads Show all messages Hide all messages | | easy dfs | 👑TIMOFEY👑`~ | 1367. Top Secret | 18 Aug 2024 13:22 | 3 | | | Very easy problem. Why so few people got ac? | ahyangyi_newid | 1367. Top Secret | 14 Aug 2024 20:17 | 5 | 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! 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 | | if you have RE19 | ~'Yamca`~ | 1367. Top Secret | 12 Aug 2024 21:26 | 1 | | | What is the answer for this testcase? | Deepesson | 1367. Top Secret | 9 Jan 2021 02:38 | 3 | #+-+-+-+-+-+
+-+-+-+-+-+#
Edited by author 09.01.2021 01:54 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 Btw the correct answer is:
11
11 | | Precautions | Denis Koshman | 1367. Top Secret | 24 Aug 2008 05:35 | 1 | 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). | | I think the problem statement is slightly incorrect (+) | Michael Rybak (accepted@ukr.net) | 1367. Top Secret | 24 Aug 2008 04:56 | 2 | 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. Or those 4 sides are diagonal sides | | Test 2 Incorrect! | svr | 1367. Top Secret | 17 Dec 2007 15:55 | 2 | 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 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 | | What answer for... | Anatoliy 'Tolyan_NO' Tolstobroff | 1367. Top Secret | 25 Aug 2007 14:01 | 9 | #+#
+#+
#+#
and
#|#
-#-
#|#
and
#-#
+#-
#+# My AC program outputs:
10100
01100
11111
00110
00101
and
10100
01100
11111
00110
00101
and
11100
11100
11111
00110
00101
I check this test but my program get Wa21.
WHO TAKE wa21 Please help me or give some good tests.
Asked me answer for this
###
###
###
please. My AC program outputs:
110110000
111111000
011011000
110110110
111111111
011011011
000110110
000111111
000011011
thanks.
Now i got Wa22
What answer for
#-#
and fo
#
|
# I got AC I am very Happy.
I find error.
this test
##.
#..
..#
Answer is :
1111
1111
1111
1111
??? 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 | | Nice problem | Gigzzz(gigz@inbox.ru) | 1367. Top Secret | 13 Aug 2005 00:55 | 8 | Realy fun problem:)
Memory/Time limit ballance
interesting, somebody can solve it without using STL(i mean not using something like deque)?
<%
print('sdf');
%>
Edited by author 05.05.2005 13:15 What you mean by STL? Are not you solve it by filling in a square table? Memory limit (in my opinion) is sufficiently loose. STL - standart template library
I fill square table, but i have problem with memory to store queue. 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. 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). I think many people always solve problems without STL. |
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