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
I figured out the test #3, it has T = 2, T = T = 1, T = 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?
The answer is 4. Because collect '4' needs 5 steps, but collect '1' needs only 4 steps.
You're wrong. The answer is equal for every colour cause you cann't collect only one colour but all the colours together. 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 (