HDU - 5983 Pocket Cube

最新推荐文章于 2018-10-08 12:01:37 发布

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最新推荐文章于 2018-10-08 12:01:37 发布

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本文链接：https://blog.youkuaiyun.com/zbspy_ZJF/article/details/78258370

本文介绍了一个关于2x2魔方（Pocket Cube）的编程问题，探讨如何通过一步操作来判断魔方是否可以被还原。文章提供了一段C++代码实现，该代码通过枚举所有可能的一次转动方式来检查魔方的状态。

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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.

Input

The 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 pieces
labelled 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 are
given 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 are
given 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 are
given 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 given
corresponding 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 given
corresponding 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 development
as 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 |
        + - + - +

Output

For each test case, output YES if can be restored in one step, otherwise output NO.

Sample Input

Sample Output

YES
YES
YES
NO

题意，给出一个2阶魔方，输入的24个数是按照图上a-x输入的，问一步之内是否能还原魔方.

说出来你可能不信，比赛的时候我为了做这个题，还自己做了个2阶魔方，根据题目给的图将24块按照a-x编号为1-24；

然后就发现一共就6种转法，然后暴力枚举6种转法就ac了23333333.

#include <iostream>

using namespace std;
int a[30], b[30];

int judge()
{
	int i, j;
	for(i = 1;i <= 24;i += 4)
	{
		for(j = i + 1;j < i + 4;j++)
		{
			if(b[j] != b[j - 1])
				return 0;
		}
	}
	return 1;
}

void cpy()
{
	for(int i = 1;i <= 24;i++)
		b[i] = a[i];
}

int main()
{
	ios::sync_with_stdio(false);
	int t;
	int i, flag;
	cin>>t;
	while(t--)
	{
		for(i = 1;i <= 24;i++)
		{
			cin>>a[i];
			b[i] = a[i];
		}
		flag = judge();
		if(!flag)
		{
			b[1] = b[5];
			b[3] = b[7];
			b[5] = b[9];
			b[7] = b[11];
			b[9] = b[13];
			b[11] = b[15];
			b[13] = a[1];
			b[15] = a[3];
			flag = judge();
			if(!flag)
			{
				cpy();
				b[15] = b[11];
				b[13] = b[9];
				b[11] = b[7];
				b[9] = b[5];
				b[7] = b[3];
				b[5] = b[1];
				b[3] = a[15];
				b[1] = a[13];
				flag = judge();
			}
		}
		if(!flag)
		{
			cpy();
			b[1] = b[21];
			b[2] = b[22];
			b[21] = b[12];
			b[22] = b[11];
			b[12] = b[17];
			b[11] = b[18];
			b[17] = a[1];
			b[18] = a[2];
			flag = judge();
			if(!flag)
			{
				cpy();
				b[18] = b[11];
				b[17] = b[12];
				b[11] = b[22];
				b[12] = b[21];
				b[22] = b[2];
				b[21] = b[1];
				b[2] = a[18];
				b[1] = a[17];
				flag = judge();
			}
		}
		if(!flag)
		{
			cpy();
			b[17] = b[7];
			b[19] = b[8];
			b[7] = b[24];
			b[8] = b[22];
			b[24] = b[14];
			b[22] = b[13];
			b[14] = a[17];
			b[13] = a[19];
			flag = judge();
			if(!flag)
			{
				cpy();
				b[13] = b[22];
				b[14] = b[24];
				b[22] = b[8];
				b[24] = b[7];
				b[8] = b[19];
				b[7] = b[17];
				b[19] = a[13];
				b[17] = a[14];
				flag = judge();
			}
		}
		if(flag)
			cout<<"YES"<<endl;
		else
			cout<<"NO"<<endl;
	}
	return 0;
}