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Discussion of Problem 1878. Rubinchik's CubeACSpeed Can anyone explain to me the input ? [1] // Problem 1878. Rubinchik's Cube 24 Nov 2011 14:03 I don't understand how to translate them to the initial state . Tks Yuv Re: Can anyone explain to me the input ? // Problem 1878. Rubinchik's Cube 17 Mar 2024 22:20 This is 4 layered Rubics Cube. These layers are situated behind each other (perpendicularly). You can think them as 4 Cards placed behind each other. Also, each layer is transparent. So, color on the second, third or fourth layer can be seen through the first layer. Top View is watching the cube from the first layer. As all the layers are transparent, so we can see the colors of each layer behind it from the first layer. So, from the TOP VIEW you will see a complete cube combining each layer's colors. In the description, the first picture is the 4 layers of the cube. Can you place them behind each other? What will you see if you observe them from the first layer and considering them as transparent? If you see them like below, then bingo you get the idea!! YRYB RRYY GGBB RGBG Here, the first Y and the last B color of the first row are coming from the 4th layer if you see. And so on for the rests. In the given input, let say the color code is like below, 1 - Red 2 - Yellow 3 - Blue 4 - Green. So, the input basically says that what you will see all the layers from the front of the first layer. If you replaced the number with the color, then you will get the exact same cube that is given in the description. (We have already built it above!!) morbidel WA3 [17] // Problem 1878. Rubinchik's Cube 22 Oct 2011 16:05 Any hint for WA3? Anil Kishore Re: WA3 [15] // Problem 1878. Rubinchik's Cube 22 Oct 2011 16:21 'm stuck with WA5 :( ZARHANE Re: WA3 // Problem 1878. Rubinchik's Cube 22 Oct 2011 17:41 I have WA40 ! Strekalovsky Oleg [Vologda SPU #1] Re: WA3 [13] // Problem 1878. Rubinchik's Cube 23 Oct 2011 23:43 Anil Kishore wrote 22 October 2011 16:21 'm stuck with WA5 :( Don't forget rotate clockwise and anticlockwise at one step!morbidel Re: WA3 [12] // Problem 1878. Rubinchik's Cube 24 Oct 2011 16:20 Yes, rotating just 90 degrees in either directions takes one step. What's the trick here? MOPDOBOPOT (USU) Re: WA3 [11] // Problem 1878. Rubinchik's Cube 25 Oct 2011 13:49 This test helped me to overcome WA3: 4 2 2 3 3 1 4 1 3 2 3 1 1 4 4 2 The answer is 4. Because collect '4' needs 5 steps, but collect '1' needs only 4 steps. morbidel Re: WA3 [7] // Problem 1878. Rubinchik's Cube 25 Oct 2011 15:15 My answer is 4. I made a classic backtracking which gets WA3 and also a Greedy who also gets WA3, which works as follows: - assume the square is solved with 1 in the upper left corner, compute the orientation of each of the layers (0 if ok, 1 or 3 if with a rotation we reach 1 in the corner, 2 if with 2 rotations). - now try to rotate the layers to achieve the color i in the upper left corner - so for each layer determine the minimal nr. of steps to move the color i from layer j in its correct position - compute the minimal number of moves to reach the solution with color i in upper-left - print the minimum out of those values Edited by author 25.10.2011 15:22 morbidel Re: WA3 [6] // Problem 1878. Rubinchik's Cube 26 Oct 2011 00:00 I figured out the test #3, it has T[0][0] = 2, T[0][1] = T[1][0] = 1, T[1][1] = 3, "if (...) while(1);" did the trick. So the whole table I assume is 2 1 2 4 1 3 1 2 3 4 2 4 1 3 4 3 My program outputs 2 as a minimum (rotate layer 4 clockwise with 90, rotate layer 3 anti-clockwise with 90), resulting a solved cube with 1 in its left corner. The judge answer is 3, but I don't understand why. Am I missing something? Edited by author 26.10.2011 00:04 morbidel Re: WA3 // Problem 1878. Rubinchik's Cube 27 Oct 2011 01:13 AC at last! The problem in my solution was that I assumed always that the colors were 1-red, 2-yellow, 3-green, 4-blue, but they say that one number is one color so they could be permuted in any way. Edited by author 27.10.2011 01:55 Lucian Ilea Re: WA3 [4] // Problem 1878. Rubinchik's Cube 7 Nov 2011 21:37 My AC program gives 2 as the answer for: 2 1 2 4 1 3 1 2 3 4 2 4 1 3 4 3 Soucup Adrian Re: WA3 [3] // Problem 1878. Rubinchik's Cube 29 Jun 2012 20:52 Lucian Ilea wrote 7 November 2011 21:37 My AC program gives 2 as the answer for: 2 1 2 4 1 3 1 2 3 4 2 4 1 3 4 3 Strange. Mine gives 3. (Also AC). Edited by author 29.06.2012 20:52 Squire [Lviv NU] Re: WA3 // Problem 1878. Rubinchik's Cube 29 Aug 2012 23:21 Soucup Adrian wrote 29 June 2012 20:52 Lucian Ilea wrote 7 November 2011 21:37 My AC program gives 2 as the answer for: 2 1 2 4 1 3 1 2 3 4 2 4 1 3 4 3 Strange. Mine gives 3. (Also AC). Edited by author 29.06.2012 20:52 tepamid Re: WA3 [1] // Problem 1878. Rubinchik's Cube 2 Jul 2019 19:32 Mine gives 2 and I got AC for this input: 2 1 2 4 1 3 1 2 3 4 2 4 1 3 4 3 Edited by author 02.07.2019 19:33 maslowmw Re: WA3 // Problem 1878. Rubinchik's Cube 17 May 2020 00:33 But this test incorrect, because always colours order 1-2-3-4 IN this tet 1-2-4-3. My AC solution also gives 3. Squire [Lviv NU] Re: WA3 [2] // Problem 1878. Rubinchik's Cube 29 Aug 2012 23:31 MOPDOBOPOT (USU) wrote 25 October 2011 13:49 This test helped me to overcome WA3: You're wrong. The answer is equal for every colour cause you cann't collect only one colour but all the colours together.4 2 2 3 3 1 4 1 3 2 3 1 1 4 4 2 The answer is 4. Because collect '4' needs 5 steps, but collect '1' needs only 4 steps. To collect '4' you must turn left inner square and turn right two outer squares each for 1 time. So we have 3 steps in sum. P. S. Sorry for bad english ( Edited by author 29.08.2012 23:31 Edited by author 29.08.2012 23:31 Edited by author 29.08.2012 23:33 MOPDOBOPOT (USU) Re: WA3 [1] // Problem 1878. Rubinchik's Cube 30 Aug 2012 00:09 If we turn squares as you said we will get this situation: 4 3 3 3 4 4 3 2 4 1 2 2 1 1 1 2 Squire [Lviv NU] Re: WA3 // Problem 1878. Rubinchik's Cube 30 Aug 2012 01:16 MOPDOBOPOT (USU) wrote 30 August 2012 00:09 If we turn squares as you said we will get this situation: Yes, you're right )))4 3 3 3 4 4 3 2 4 1 2 2 1 1 1 2 We need one more step - turn right the square which includes element table[3][1]. So, the answer is 4. inatial_D Re: WA3 // Problem 1878. Rubinchik's Cube 5 Mar 2013 16:55 ac is reached just using 4 elements in inputs. RinchinGaY ez problem c++ accept too easy lolololololllol solved 1 sec // Problem 1878. Rubinchik's Cube 20 Mar 2019 19:16 #include <iostream> #include <vector> #include <cmath> #include <algorithm> #include <set> #include <unordered_map> #include <map> #include <string> using namespace std; int main() { vector<vector<int>> v(4, vector<int>(4)); vector<int> col(4); for(int i= 0 ; i < 4; i++) { for(int j = 0; j < 4; j++) { cin >> v[i][j]; if(v[i][j] == 1) { if(((i == 0 || i == 1) && ((j == 0) || (j == 1)))) { col[0]++; } else if((i == 0 || i == 1) && ((j == 2) || (j == 3))) { col[1]++; } else if((i == 2 || i == 3) && (j == 0 || j == 1)) { col[2]++; } else { col[3]++; } } } } cout << min(min(col[1] + col[2] + col[3] + col[3], col[2] + col[0] + col[2] + col[3]), min(col[1] + col[2] + col[0] + col[0], col[0] + col[3] + col[1] + col[1])); } Kirill can anyone explain me how to do it??? // Problem 1878. Rubinchik's Cube 21 Feb 2018 15:08 explain me the way of doing it or give a solution Edited by author 21.02.2018 16:21 Filip Franik Tests for WA7 // Problem 1878. Rubinchik's Cube 14 Aug 2017 19:48 3 1 2 4 1 3 4 2 4 2 1 3 2 4 3 1 (4) 4 1 2 1 1 4 1 2 4 3 2 3 3 4 3 2 (2) Morteza Balvardi wa #21 // Problem 1878. Rubinchik's Cube 10 Mar 2014 18:43 can anybody provide a test case for this error my method is quiet like a brute force but it seems i have forgotten sth Edited by author 10.03.2014 18:44 R. Dubinin WA # 7 // Problem 1878. Rubinchik's Cube 7 Aug 2013 17:30 please give me some tests Edited by author 07.08.2013 17:31 Yagudin Ruslan WA 37 // Problem 1878. Rubinchik's Cube 22 Oct 2012 10:38 Does correct test? Please give me test. Anatoly AC approach [3] // Problem 1878. Rubinchik's Cube 21 Feb 2012 23:11 Brute force algo is just 4^5 actions. So, you can don't think about logical solution :) Andrew Sboev Re: AC approach [2] // Problem 1878. Rubinchik's Cube 30 May 2012 19:46 He-he, I could not understand how to solve it by bruteforce algo, because my bruteforce doesnt't pass some tests :) Logical algo is very simple. Edited by author 02.06.2012 15:45 Soucup Adrian Re: AC approach [1] // Problem 1878. Rubinchik's Cube 29 Jun 2012 20:50 Brute force algo can be done in 2^4 steps and works. Andrew Sboev Re: AC approach // Problem 1878. Rubinchik's Cube 30 Jun 2012 21:03 Hmm, my logical algo takes also 16 operations. Four iterations of cycle with four operations in each iteration. Edited by author 30.06.2012 21:04
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