HDU 5292 Pocket Cube

Pocket Cube

Problem Description

Pocket Cube is the 2×2×2 equivalent of a Rubik’s Cube(3×3×3). The cube consists of 8 pieces, all corners. (from wiki)

 

The right position rotates in red face clockwisely. 
You can get more details from input case.
w represents white , y represents yellow , o represents orange , r represents red , g represents green , b represents blue. In the right position, white and yellow , orange and red , green and blue are in the opposite face. 

 

Input

The first line of input contains only one integer T(<=10000), the number of test cases. 
Each case contains a Pocket Cube described above. After each case , there is a blacnk line. It guarantees that the corners of the Cube is right.

 

Output

Each case contains only one line. Each line should start with “Case #i: ”,with i implying the case number, followed by “YES” or “NO”,”YES” means you can return it to the right position, otherwise “NO”.

 

Sample Input

2
    g y 
    g y 
o o w g r r 
o o w g r r 
    b w 
    b w 
    y b 
    y b

    r w 
    g b 
b y o w o r 
y g y o g b 
    r w 
    o y 
    b r 
    w g

 

Sample Output

Case #1: YES
Case #2: NO

/*
 * 二阶魔方任何一个块都可以旋转，这里考虑对色面，（黄，白）， （绿， 蓝）， （红， 橙）
 * 将标准魔方展开
 *          0  0
 *          0  0
 *    1 -1  1  -1  1  -1
 *    -1  1  -1  1  -1  1
 *          0  0
 *          0  0
 *          1  -1
 *          -1  1
 * 只要考虑一组对立面即可，比如（黄， 白），对应到上述展开后，只要最后总和模三等于零就是合理的
 */

#include <bits/stdc++.h>
#define endl "\n"
using namespace std;

const int cube[] = {0, 0, 0, 0, 1, -1, 1, -1, 1, -1, -1, 1, -1, 1, -1, 1, 0, 0, 0, 0, 1, -1, -1, 1};

int main() {
    int T;
    cin >> T;
    for(int cas = 1; cas <= T; ++cas) {
        char color[2];
        int ret = 0;
        for(int i = 0; i < 24; ++i) {
            scanf("%s", color);
            if(color[0] == 'w' || color[0] == 'y') ret += cube[i];
        }
        cout <<  "Case #" << cas << ": " << (ret % 3 ? "NO" : "YES") << endl;
    }
    return 0;
}