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 ab,r, org. The character ( ) from the left gives the color of facei. For example, Figure 2 is a picture ofrbgggr and Figure 3 corresponds torggbgr. 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

rbgggrrggbgrrrrbbbrrbbbrrbgrbgrrrrrg

Sample Output

TRUEFALSEFALSE

     其实很简单，但刚开始看题目想了很久纠结怎么确定是不是同一个正方体。。。。上网看解题报告才明白，.正方体是对称的，在第一个正方体任意选三个面，每个面试图都在第二个正方体中找到与这个面相同且相应的对面也相同，如果找不到就直接false，找到了就对第二个正方体进行标记防止同一个面重复使用。

#include <iostream>using namespace std;int main(){     char a[13];     char b[7];     while ( cin >> a)     {         for( int i = 6; i<12; i++)              b[i-6] = a[i];         b[6] = 0;         int ans = 0;         for( int i = 0; i < 3; i++){             ans = 0;             for( int j = 0; j < 6; j++){             if( a[i] == b[j] && a[5-i] == b[5-j])             {                 ans = 1; b[j] = 'a'; b[5-j]='a';                  break;             }         }            if ( !ans){                cout << "FALSE" << endl;                break;            }         }         if( ans)         cout << "TRUE" << endl;     }    return 0;}