UVA 253 Cube painting

Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu
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Description
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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.

picture21

Figure 1.

Since a cube has 6 faces, our machine can paint a face-numbered cube in tex2html_wrap_inline126 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 tex2html_wrap_inline128 character ( tex2html_wrap_inline130 ) 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

/*
Author:ZCC;
Time:2015-6-4
Solve:分析可知1-6，2-5，3-4，所以两个骰子若相同的话，必定面对面的花色是一样的
所以只要判断两个骰子的所形成的三个对面是否都能够对应起来就可以了
*/

#include<iostream>
#include<algorithm>
#include<map>
#include<cstdio>
#include<cstdlib>
#include<vector>
#include<cmath>
#include<cstring>
#include<string>
using namespace std;
const int maxn=13;
typedef long long LL;
char str[maxn];
char s1[200][200],s2[200][200];
int main(){
    #ifndef ONLINE_JUDGE
    freopen("Text//in.txt","r",stdin);
    #endif // ONLINE_JUDGE
    while(~scanf("%s",str)){
        memset(s1,0,sizeof(s1));
        memset(s2,0,sizeof(s2));
        for(int i=0;i<=2;i++){
            s1[str[i]][str[5-i]]++;
            s1[str[5-i]][str[i]]++;
        }
        for(int i=6;i<=8;i++){
            s2[str[i]][str[12-i+5]]++;
            s2[str[12-i+5]][str[i]]++;
        }
        int i;
        for( i=0;i<3;i++){
            if(s1[str[i]][str[5-i]]!=s2[str[i]][str[5-i]])break;
        }
        if(i>=3)puts("TRUE");
        else puts("FALSE");
    }
    return 0;
}