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

要注意涂色的顺序是按题目中给的

上前左右后下

 1  2  3  4  5  6
 1  4  2  5  3  6
 1  5  4  3  2  6
 1  3  5  2  4  6
 6  2  4  3  5  1
 ...
 2  3  1  6  4  5
 ...
 5  4  1  6  3  2
 ....
 3  2  6  1  5  4
 ...
 4  2  1  6  5  3
 up   1
down 6
a    2
b    3
c    4
d    5

up dwon 有六种组合其对应的abcd有四种共24种，

 以上下为轴，左右为轴，前后为轴
每个8种，上下可以翻转。
其余四个面顺时针旋转：
前-》左
左-》后
后-》右
右-》前3    4   5   6

没有想到简单的表示方法只能手打穷举了......

#include
#include
char s[12];
int sort(char up,char down,char a,char b,char c,char d)
{if ((up==s[6])&&(down==s[11])&&(a==s[7])&&(b==s[8])&&(c==s[9])&&(d==s[10])) return 1;
 if ((up==s[6])&&(down==s[11])&&(c==s[7])&&(a==s[8])&&(d==s[9])&&(b==s[10])) return 1;
 if ((up==s[6])&&(down==s[11])&&(d==s[7])&&(c==s[8])&&(b==s[9])&&(a==s[10])) return 1;
 if ((up==s[6])&&(down==s[11])&&(b==s[7])&&(d==s[8])&&(a==s[9])&&(c==s[10])) return 1;
 return 0;
};
void main()
{int f1,f2,f3,f4,f5,f6;
 while (scanf("%s",&s)!=EOF)
 {f1=sort(s[0],s[5],s[1],s[2],s[3],s[4]);
  f2=sort(s[1],s[4],s[2],s[0],s[5],s[3]);
  f3=sort(s[2],s[3],s[1],s[5],s[0],s[4]);
  f4=sort(s[5],s[0],s[1],s[3],s[2],s[4]);
  f5=sort(s[4],s[1],s[3],s[0],s[5],s[2]);
  f6=sort(s[3],s[2],s[1],s[0],s[5],s[4]);
  if (f1+f2+f3+f4+f5+f6) printf("TRUE\n");
  else printf("FALSE\n");
 }
}