大體題意：

給你一個2*2*2的魔方，問你能否一步到達各個面的顏色完全一樣。

思路：

其實挺簡單的，題意已經(jīng)給足了提示，已經(jīng)告訴你了魔方怎么進行標號。（就是那個圖）

只要給魔方標號，怎么轉(zhuǎn)就很簡單了，可以預(yù)處理一個b 數(shù)組和c 數(shù)組，分別是魔方轉(zhuǎn)一個面時候的側(cè)面的八個面和上面的四個面，循環(huán)賦值即可。

這樣 寫好循環(huán)賦值函數(shù)后，轉(zhuǎn)六個面只需要更改b數(shù)組和c數(shù)組即可。

詳細見代碼：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int a[25];bool ok;int ori[25];void fuyuan(){    for (int i = 0; i < 24; ++i) a[i] = ori[i];}bool check(){    for (int i = 0; i < 24; i += 4){        for (int j = i; j < i+4; ++j){            if (a[j] != a[i]) return false;        }    }    return true;}int b[] = {14,15,20,22,5,4,19,17};int c[] = {2,3,1,0};void zhuan(){    int t1 = a[b[0]], t2 = a[b[1]];    for (int i = 0; i < 6; i += 2){        int id1 = b[i], id2 = b[i+1];        int nid1 = b[i+2], nid2 = b[i+3];        a[id1] = a[nid1];        a[id2] = a[nid2];    }    a[b[6] ] = t1; a[b[7] ] = t2;    if (check()) ok = 1;fuyuan();    t1 = a[b[7]], t2 = a[b[6]];    for (int i = 7; i > 2; i -= 2){        int id1 = b[i], id2 = b[i-1];        int nid1 = b[i-2], nid2 = b[i-3];        a[id1] = a[nid1];        a[id2] = a[nid2];    }    a[b[0] ] = t2; a[b[1] ] = t1;    if (check()) ok = 1; fuyuan();}void add(int a0,int a1,int a2,int a3,int a4,int a5,int a6,int a7,int c0,int c1,int c2,int c3){    b[0] = a0;b[1] = a1;b[2] = a2;b[3] = a3;b[4] = a4;b[5] = a5;b[6] = a6;b[7] = a7;    c[0] = c0;c[1] = c1;c[2] = c2;c[3] = c3;}void up(){    add(14,15,20,22,5,4,19,17,2,3,1,0);    zhuan();}void down(){    add(12,13,21,23,7,6,18,16,8,9,11,10);    zhuan();}void Left(){    add(2,0,14,12,10,8,6,4,19,17,18,16);    zhuan();}void Right(){    add(3,1,15,13,11,9,7,5,20,22,23,21);    zhuan();}void qian(){    add(0,1,20,21,11,10,16,17,14,15,13,12);    zhuan();}void hou(){    add(2,3,22,23,9,8,18,19,4,5,6,7);    zhuan();}int main(){    int T;    scanf("%d",&T);    while(T--){        ok = 0;        for (int i = 0; i < 24; ++i) scanf("%d",a+i);        for (int i = 0; i < 24; ++i)ori[i] = a[i];        if (check()) ok = 1;        up(); down(); Left(); Right(); qian(); hou();        if (ok)puts("YES");        else puts("NO");    }    return 0;}
Pocket Cube
Time Limit: 2000/1000 MS (java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 358    Accepted Submission(s): 131PRoblem DescriptionThe Pocket Cube, also known as the Mini Cube or the Ice Cube, is the 2 × 2 × 2 equivalence of a Rubik’s Cube.The cube consists of 8 pieces, all corners.Each piece is labeled by a three dimensional coordinate (h, k, l) where h, k, l ∈ {0, 1}. Each of the six faces owns four small faces filled with a positive integer.For each step, you can choose a certain face and turn the face ninety degrees clockwise or counterclockwise.You should judge that if one can restore the pocket cube in one step. We say a pocket cube has been restored if each face owns four same integers. InputThe first line of input contains one integer N(N ≤ 30) which is the number of test cases.For each test case, the first line describes the top face of the pocket cube, which is the common 2 × 2 face of pieceslabelled by (0, 0, 1),(0, 1, 1),(1, 0, 1),(1, 1, 1). Four integers are given corresponding to the above pieces.The second line describes the front face, the common face of (1, 0, 1),(1, 1, 1),(1, 0, 0),(1, 1, 0). Four integers aregiven corresponding to the above pieces.The third line describes the bottom face, the common face of (1, 0, 0),(1, 1, 0),(0, 0, 0),(0, 1, 0). Four integers aregiven corresponding to the above pieces.The fourth line describes the back face, the common face of (0, 0, 0),(0, 1, 0),(0, 0, 1),(0, 1, 1). Four integers aregiven corresponding to the above pieces.The fifth line describes the left face, the common face of (0, 0, 0),(0, 0, 1),(1, 0, 0),(1, 0, 1). Four integers are givencorresponding to the above pieces.The six line describes the right face, the common face of (0, 1, 1),(0, 1, 0),(1, 1, 1),(1, 1, 0). Four integers are givencorresponding to the above pieces.In other Words, each test case contains 24 integers a, b, c to x. You can flat the surface to get the surface developmentas follows.
+ - + - + - + - + - + - +| q | r | a | b | u | v |+ - + - + - + - + - + - +| s | t | c | d | w | x |+ - + - + - + - + - + - +        | e | f |        + - + - +        | g | h |        + - + - +        | i | j |        + - + - +        | k | l |        + - + - +        | m | n |        + - + - +        | o | p |        + - + - + OutputFor each test case, output YES if can be restored in one step, otherwise output NO. Sample Input
41 1 1 12 2 2 23 3 3 34 4 4 45 5 5 56 6 6 66 6 6 61 1 1 12 2 2 23 3 3 35 5 5 54 4 4 41 4 1 42 1 2 13 2 3 24 3 4 35 5 5 56 6 6 61 3 1 32 4 2 43 1 3 14 2 4 25 5 5 56 6 6 6 Sample Output
YESYESYESNO Source2016ACM/ICPC亞洲區(qū)青島站-重現(xiàn)賽（感謝中國石油大學(xué)） Recommendjiangzijing2015   |   We have carefully selected several similar problems for you:  6014 6013 6012 6011 6010