| Show all threads Hide all threads Show all messages Hide all messages | | Hint | yyll | 1509. Domino Recognition | 14 Jun 2021 17:23 | 1 | Hint yyll 14 Jun 2021 17:23 Find a robust feature to distinguish different bones with same total dots.
This really is an easy problem once you find that feature. | | help with 52 pls | 🎧 Vadim Barinov \Frez_Fstilus/'` | 1509. Domino Recognition | 14 Jan 2018 02:11 | 2 | What may cause this judgement result? I think there is an error in your domino size cheking function or something similar. See your other topic for details. | | Sol with WA 46, 47, 48 | 🎧 Vadim Barinov \Frez_Fstilus/'` | 1509. Domino Recognition | 14 Jan 2018 02:08 | 2 | WA 46:
2
0 0
0 0.5
Ans:
0 2
WA 47:
2
0 0
0 99
Ans:
1 1
WA 48:
2
0 0
0 1
Ans:
0 2
1 1 > WA 46:
> 2
> 0 0
> 0 0.5
> Ans:
> 0 2
Your answer is incorrect. There is no dominoes which fit with your input. Maximum related L (for 0-2 domino) is 1/sqrt(2) that is outside of permissible range of [1, 100].
But thank you nonetheless!!! Your hint allowed me to get my AC 8-] | | Самая непонятная задача!!!!! | xurshid_n | 1509. Domino Recognition | 21 Aug 2012 15:51 | 1 | В условии написано, что домино перевёрнута.
Но я отыслал перевёрнутая версия - ошибка, а не перевёрнутая версия принита, что за задача?
Edited by author 21.08.2012 15:51
Edited by author 21.08.2012 16:22 | | One Hint and a lot of tests | Leonid (SLenik) Andrievskiy | 1509. Domino Recognition | 1 May 2011 02:46 | 1 | 1. If you get WA1 - check if you considered that the domino is mirrored.
2. Some tests (e.g. training samples), which is grouped by the number of points:
0
1
2 2
2
2 2
6 2
2
1 1
3 3
3
1 1
3 3
6 2
3
1 1
2 2
3 3
4
5 1
5 3
7 1
7 3
4
1 1
2 2
3 3
6 2
4
1 1
3 3
5 1
7 3
5
5 1
5 3
6 2
7 1
7 3
5
6 2
1 1
1 3
3 1
3 3
5
1 1
2 2
3 3
5 1
7 3
6
5 1
5 3
6 1
6 3
7 1
7 3
6
6 2
1 1
1 3
3 1
3 3
2 2
6
1 1
1 3
3 1
3 3
5 1
7 3
6
1 1
2 2
3 3
5 1
6 2
7 3
7
6 2
1 1
2 1
3 1
1 3
2 3
3 3
7
1 1
2 2
1 3
3 1
3 3
5 1
7 3
7
5 1
6 2
7 3
1 1
1 3
3 1
3 3
8
1 1
2 1
3 1
1 3
2 3
3 3
5 1
7 3
8
1 1
1 3
2 2
3 1
3 3
5 1
6 2
7 3
8
1 1
1 3
3 1
3 3
5 1
5 3
7 1
7 3
9
1 1
2 2
1 3
3 1
3 3
5 1
5 3
7 1
7 3
9
5 3
6 2
7 1
1 1
2 1
3 1
1 3
2 3
3 3
10
1 1
1 3
2 2
3 1
3 3
5 1
5 3
6 2
7 1
7 3
10
1 1
1 3
2 1
2 3
3 1
3 3
5 1
5 3
7 1
7 3
11
5 1
5 3
6 2
7 1
7 3
1 1
1 3
2 1
2 3
3 1
3 3
12
1 1
2 1
3 1
1 3
2 3
3 3
5 1
5 3
6 1
6 3
7 1
7 3 | | It seems that test 51 has garbage after test data | hedrok <ONPU> | 1509. Domino Recognition | 2 Jun 2010 18:48 | 2 | I read input using
while (scanf("%d", &n) != EOF)
{
...
}
to be able to run multiple tests. I had TLE 51 with such code.
But when i deleted this cycle, i got accepted. Please check test 51. Test 51 is correct. You have got TLE 51 because of bug in your code. | | Что за тесты? | AndreySUrSU | 1509. Domino Recognition | 31 Aug 2008 08:59 | 9 | Я из команды SUrSU 1.
Не понятно, что у вас за тесты на эту задачу.
Говорится, что координаты даны с точностью до 5 знаков, мы предполагали, что достаточно точно.
Посылали с большой точностью на контесте, это было в самом конце контеста, как-то не успели подумать, что можно и с маленькой точностью послать, к тому же еще были заняты написанием задачи И.
Теперь я написал такое же решение, посылал с точностью 10^-6 вроде - не принималась, 10^-2 - принялась.
Что за бред???? Мда, это все конечно интересно, но мое решение например вообще нигде не использует точность и AC c первого раза. Но правда и задача очень специфическая. Все верно, если входные данные зашумлены на 1E-5, то требовать от решения 1E-6 неверно Это у нас используется для проверки коэффициента подобия.
1E-4 - 8WA
1E-3 - 44WA
Все-таки странно.
Почему именно 1E-2 проходит, а 1E-3 - плохая точность. Посчитаем погрешность вычисления коэффициента подобия: погрешность координат 1e-5, значит, погрешность расстояния — 3e-5, погрешность коэффициента при максимальном стократном отношении размеров — 3e-3.
Понятно, что точности 1e-3 недостаточно для сравнения коэффициентов подобия.
Edited by author 01.11.2006 17:41 Что-то я не совсем понимаю это.
У нас есть эталонная доминошка со стороной 1. Координаты на ней точные. 1<=K<=100. Любые доминошки, с меньшими расстояниями не подходят.
Рассматриваем доминошки с большими расстояниями, они вроде бы, по логике, должны быть более точно засняты. Получается, что чем крупнее снимок, тем хуже точность, странно.
По-моему у крупного снимка относительная точность больше, те с ростом коэффициента точность растет.
Или я не прав? I tried to follow the topic but the translator russian - english was not enough. I have a program which gets WA in test 44. Actually I proved changing precision issues and the best I got was WA in test 90, so I think the only bug in the code is about precision.
The function that deals with precision in my code is:
bool check(par *q, double t) {
int i, j, orden[12];
bool usado[12];
memset(usado, 0, sizeof(usado));
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) if (!usado[j])
if (dist(q[i].first - p[j].first, q[i].second - p[j].second) < eps) break;
if (j == n) return false;
usado[j] = true;
}
return true;
}
Array p keeps the input points and array q the points in a possible domino tile. The function check returns if the points in both arrays are "more or less" the same
The function dist is:
double dist(double x, double y) {
return fabs(x)+ fabs(y);
}
and the eps is defined to be 1e-3
It would be great if anybody who got AC explains how he dealt with precision in his code This is not precision problem, but wrong approach problem.
Input coordinates do not define points. They define a SET of points whose precise coordinates become input values after rounding. E.g. pair (1.2345;6.7890) defines half-open square [1.23445;1.23455) x [6.78895;6.78905). All points within this square become (1.2345;6.7890) after rounding.
You task is to find all dies with 1<=L<=100 whose precise dot coordinates fall into these half-open squares with one-to-one relationship. | | Give me PLEASE test 46 (I think it is n=2) | Hayk | 1509. Domino Recognition | 5 Oct 2007 21:07 | 8 | I can't take this no more!!!!!!!!!!!!
PLEASE!!!!
Somebody!
Anybody!!
Tell me, what is the test#46
Or maybe you can say(and explain) Is the answer always
0 2
1 1
when n==2; For n==2 this answer is not always right. Check 1 ≤ L ≤ 100 for cases 0 2 and 1 1.
Edited by author 12.11.2006 01:39 Yes, I have checked.
if n==2:
1 1 always is correct answer, but
0 2 is correct when the distance between that two points is less or equal than 50*sqrt(2.0);
is it right?
P.S.
And the other trivial cases are:
n=1 : 0 1
n=11 : 5 6
n=12 : 6 6 1 1 is not always correct. For test:
2
0 0
0 0.9
Аnswer is 0 2 only.
А волшебство тут не причем. Просто так получилось что я сравниваю только числа, которые либо заведомо отличаются на величину много большую точности (и она никак не может повлиять на результат сравнения), либо целые числа. Друг, СПАСИБО ТЕБЕ БОЛЬШОЕ!!!!!!
Очень, очень ...n раз.... очень тебе благадарен!
(здесь n-> {почти бесконечности} ) Nechaev Ilya, thank you very much for case n=2! Ya viju u tebya AC s pervogo raza i bez proverki na tochnost'?!?!?!
Da ti mojno skazat' volshebnik.
Budu blogadaren, esli otkroesh tvoyu taynu.:)
P.S. Pogovorim na radnom yazike:) | | as | Rostislav | 1509. Domino Recognition | 28 Jan 2007 22:20 | 9 | as Rostislav 26 Jan 2007 23:41 I have reached the limit.
I tested my program with more than 10 000 tests, and did ~100 submissions but I can't understand my mistake.
Can someone give me some tricky tests (test 57 will do the work, too :) ).
Thank you.
Rostislav
P.S. my mail is (my nick)_everest(at)yahoo(dot)com
Edited by author 26.01.2007 23:46 You should recognize the difference between 0:2 and 1:1. See, 1 ≤ L ≤ 100, so if the dots are too close or too far, there is only one option (0:2 for too close, and 1:1 for too far).
P.S. I have not submitted the problem yet, so I might mistaken ;)
Edited by author 27.01.2007 00:00 Unfortunately, this is not the case :( 0:2 1:1 Differce is test Nr 43 or 46, if remember well. Well, if you want I could send you my AC program. I will be greatful!
(my e-mail is rostislav_everest (_at) yahoo (_dot) com)
Rostislav 5
0.4764 1.5077
0.2303 1.2538
0.7224 1.7616
0.7303 1.2616
0.2225 1.7537
Can someone with accepted solution tell me the answer.
I believe it is 0 5. Finally! Thank you for your cooperation.
Rostislav | | 1509 | svr | 1509. Domino Recognition | 21 Jan 2007 23:29 | 5 | 1509 svr 5 Dec 2006 11:23 Problem really simple.
Use harsh criterion of koeff between max and min distances
stable for rounding ang if N=7 also number of pairs with min koeff.
Most difficult situation if N=2 and distance near
50*sqrt(2), if using rounding rooles here maybe very many tricky tests but authors kindly didn't use this opportunities. Yep, problem rather simple, but i made more then 30 submission before AC, and fully rewrited my program 3 times.
Also there are something strage with the precision. Re: 1509 Fyodor Menshikov 21 Jan 2007 15:47 Also there are something strage with the precision. Yes, there are strange things with precision. After normalizing max distance to 1, not all metrics worked: sum of absolute values of differences between corresponding distances -> ac sum of squares of differences between corresponding distances -> wa8 maximum of absolute values of differences between corresponding distances -> wa8 Re: 1509 Fyodor Menshikov 21 Jan 2007 15:55 problem rather simple, but i made more then 30 submission before AC I made exactly 30 submissions before AC. I think this number is standard for this problem. :-) I think that float numbers troubles tipical thing.
For float math lows don't satisfies and result depend on
algorithm when for right math result dosn't depend on and we can use different algos. It's humor then men play with epsilon trying to ajust to tests list. |
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