uva 253 - Cube painting

Cube painting 

We have a machine for painting cubes. It is supplied with three different colors: blue, red and green. Each face of the cube gets one of these colors. The cube's faces are numbered as in Figure 1.

 

 

 

Figure 1.

Since a cube has 6 faces, our machine can paint a face-numbered cube in  different ways. When ignoring the face-numbers, the number of different paintings is much less, because a cube can be rotated. See example below. We denote a painted cube by a string of 6 characters, where each character is a b, r, or g. The  character (  ) from the left gives the color of face i. For example, Figure 2 is a picture of rbgggr and Figure 3 corresponds to rggbgr. Notice that both cubes are painted in the same way: by rotating it around the vertical axis by 90  , the one changes into the other.

 

Input

The input of your program is a textfile that ends with the standard end-of-file marker. Each line is a string of 12 characters. The first 6 characters of this string are the representation of a painted cube, the remaining 6 characters give you the representation of another cube. Your program determines whether these two cubes are painted in the same way, that is, whether by any combination of rotations one can be turned into the other. (Reflections are not allowed.)

 

Output

The output is a file of boolean. For each line of input, output contains TRUE if the second half can be obtained from the first half by rotation as describes above, FALSE otherwise.

 

Sample Input

 

rbgggrrggbgr
rrrbbbrrbbbr
rbgrbgrrrrrg

 

Sample Output

TRUE

FALSE

FALSE

这道题目开始没想通，去看别人的代码，网上有很多人用的暴力，只看到一个很犀利的想法，如下：

设两个正方体是 c1, c2, 保持 c1 不变，枚举 c1 的相邻的三个面，和 c2 的6个面依次对比，如果这个面和其中一个面相同，并且这两个面的对面也对应相同，那么就把 c2 的这两个面标记为字符 ‘0’ ，防止以后再次用到这两个面。如果对于 c1 的这三个面的其中一个面，在 c2 中找不到符合条件（即两个面相同并且对面也相同）的，就退出循环，说明这两个正方体不相同，否则，这两个正方体相同。

唉，不愧是大牛的想法，ORZ…………

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>

using namespace std;

int main(void)
{
    char a[13], c1[6], c2[6]; 
#ifndef ONLINE_JUDGE
    freopen("in", "r", stdin);
#endif
    while (cin>>a)
    {
        int i, j = 0;
        for (i = 0; i < 6; i++)    { c1[i] = a[i]; c2[i] = a[i+6]; }
        int mrk;
        for (i = 0; i < 3; i++)
        { 
            mrk = 0;
            for (j = 0; j < 6; j++)
            {
                if (c1[i]==c2[j] && c1[5-i]==c2[5-j])
                {
                    mrk = 1; c2[j] = '0'; c2[5-j] = '0'; break;    
                }
            }
            if (!mrk)    break;
        }
        if (!mrk)    cout << "FALSE" << endl;
        else    cout << "TRUE" << endl;
    }

    return 0;
}