<h1 id="heading">题目</h1>
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The 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.  
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<p>The first line of input contains one integer N(N ≤ 30) which is the number of test cases.<br>
For each test case, the first line describes the top face of the pocket cube, which is the common 2 × 2 face of pieces<br>
labelled by (0, 0, 1),(0, 1, 1),(1, 0, 1),(1, 1, 1). Four integers are given corresponding to the above pieces.<br>
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 are<br>
given corresponding to the above pieces.<br>
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 are<br>
given corresponding to the above pieces.<br>
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 are<br>
given corresponding to the above pieces.<br>
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 given<br>
corresponding to the above pieces.<br>
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 given<br>
corresponding to the above pieces.<br>
In other words, each test case contains 24 integers a, b, c to x. You can flat the surface to get the surface development<br>
as follows.</p>
<pre><code>+ - + - + - + - + - + - +  
| q | r | a | b | u | v |  
+ - + - + - + - + - + - +  
| s | t | c | d | w | x |  
+ - + - + - + - + - + - +  
        | e | f |  
        + - + - +  
        | g | h |  
        + - + - +  
        | i | j |  
        + - + - +  
        | k | l |  
        + - + - +  
        | m | n |  
        + - + - +  
        | o | p |  
        + - + - +  
</code></pre>
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<p>For each test case, output YES if can be restored in one step, otherwise output NO.</p>
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<p>4<br>
1 1 1 1<br>
2 2 2 2<br>
3 3 3 3<br>
4 4 4 4<br>
5 5 5 5<br>
6 6 6 6<br>
6 6 6 6<br>
1 1 1 1<br>
2 2 2 2<br>
3 3 3 3<br>
5 5 5 5<br>
4 4 4 4<br>
1 4 1 4<br>
2 1 2 1<br>
3 2 3 2<br>
4 3 4 3<br>
5 5 5 5<br>
6 6 6 6<br>
1 3 1 3<br>
2 4 2 4<br>
3 1 3 1<br>
4 2 4 2<br>
5 5 5 5<br>
6 6 6 6</p>
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<p>YES<br>
YES<br>
YES<br>
NO</p>
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# 题解
题意很容易理解,就是对一个二阶魔方,判断其是否能在一步内还原.  
如果会还原三阶魔方,应该很容易就能理解.  
经过分析,其实只有7种情况  
1种是初始情况就是还原的,另外对于每一面分别向两个方向转一次.  
<p>比较稳妥的写法是直接把转后的位置顺序写出来,然后按照这个顺序查询是否存在不合法的情况.<br>
(一定要注意不要写错!!!)</p>
<p><strong>另外,需要额外注意,必须要满足6种颜色每种颜色4个</strong></p>
<h1 id="heading-1">代码</h1>
<div><div class="fold_hider"><div class="close hider_title">点击显/隐</div></div><div class="fold">```cpp Pocket Cube https://github.com/OhYee/sourcecode/tree/master/ACM 代码备份
/*//
#define debug
#include <ctime>
//*/
#include <cstdio>
#include <iostream>
#include <cstring>
#include <map>
using namespace std;
<p>int cube[24];</p>
<p>int list[7][24] = {<br>
{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23},<br>
{0,5,2,7,4,9,6,11,8,13,10,15,12,1,14,3,16,17,18,19,22,20,23,21},<br>
{0,13,2,15,4,1,6,3,8,5,10,7,12,9,14,11,16,17,18,19,21,23,20,22},<br>
{20,21,2,3,4,5,6,7,8,9,16,17,12,13,14,15,0,1,18,19,10,11,22,23},<br>
{16,17,2,3,4,5,6,7,8,9,20,21,12,13,14,15,10,11,18,19,0,1,22,23},<br>
{2,0,3,1,22,20,6,7,8,9,10,11,12,13,19,17,16,4,18,5,14,21,15,23},<br>
{1,3,0,2,17,19,6,7,8,9,10,11,12,13,20,22,16,15,18,14,4,21,5,23}<br>
};</p>
<p>int check[24]={16,17,0,1,20,21,18,19,2,3,22,23,4,5,6,7,8,9,10,11,12,13,14,15};</p>
<p>map&lt;int,int&gt; m;<br>
void solve() {<br>
m.clear();<br>
int pos = 0;<br>
int num[24];<br>
memset(num, 0, sizeof(num));</p>
<pre><code>for (int i = 0; i &lt; 6; i++){
    for (int j = 0; j &lt; 4; j++) {
        int t;
        cin &gt;&gt; t;
        cube[i * 4 + j] = t;
        if (!m.count(t))
            m.insert(make_pair(t, pos++));
        num[m[t]]++;
    }
}

// cout&lt;&lt;&quot;Cube:&quot;&lt;&lt;endl;
// for(int i=0;i&lt;24;i++)
//     cout&lt;&lt;cube[i]&lt;&lt;&quot; &quot;;
// cout&lt;&lt;endl;

//必须是6种颜色,并且每种4个

bool ok = true;
for (int i = 0; i &lt; 6 &amp;&amp; ok; i++)
    if (num[i] != 4)
        ok = false;

if (ok) {
    ok = false;

    for (int i = 0; i &lt; 7 &amp;&amp; !ok; i++) {
        //分别按照7种情况判断是否能够还原

        // cout &lt;&lt; &quot;Case&quot; &lt;&lt; i &lt;&lt; endl;
        // for(int j=0;j&lt;6;j++)
        //     printf(&quot;%3d&quot;,cube[list[i][check[j]]]);
        // cout&lt;&lt;endl;
        // for(int j=6;j&lt;12;j++)
        //     printf(&quot;%3d&quot;,cube[list[i][check[j]]]);
        // cout&lt;&lt;endl;
        // for(int j=0;j&lt;6;j++)
        //     printf(&quot;      %3d%3d\n&quot;,cube[list[i][check[12+j*2]]],cube[list[i][check[12+j*2+1]]]);
        // cout&lt;&lt;endl;
        

        bool flag = true;
        for (int j = 0; j &lt; 6 &amp;&amp; flag; j++) {
            if (cube[list[i][4 * j + 0]] != cube[list[i][4 * j + 1]] ||
                cube[list[i][4 * j + 0]] != cube[list[i][4 * j + 2]] ||
                cube[list[i][4 * j + 0]] != cube[list[i][4 * j + 3]])
                flag = false;
        }
        if (flag)
            ok = true;
    }
}
cout &lt;&lt; (ok ? &quot;YES&quot; : &quot;NO&quot;) &lt;&lt; endl;
</code></pre>
<p>}</p>
<p>int main() {<br>
#ifdef debug<br>
freopen(&quot;in.txt&quot;, &quot;r&quot;, stdin);<br>
int START = clock();<br>
#endif</p>
<pre><code>int T;
cin &gt;&gt; T;
while (T--)
    solve();
</code></pre>
<p>#ifdef debug<br>
printf(&quot;Time:%.3f s.\n&quot;, double(clock() - START) / CLOCKS_PER_SEC);<br>
#endif<br>
return 0;<br>
}</p>
<pre style="background-color:#fff">&lt;/div&gt;&lt;/div&gt;</pre>

HDU 5983.Pocket Cube(2016 CCPC 青岛 B)