UVa 253 Cube painting (模拟)

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本文深入探讨了代码逻辑的解析与优化方法，通过实例分析帮助开发者理解并提升代码效率。从基本逻辑结构到高级优化策略，全面覆盖，旨在为开发者提供实用的技术指导。

253 - Cube painting

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=99&page=show_problem&problem=189

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.

Figure 1.

Since a cube has 6 faces, our machine can paint a face-numbered cube in 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, org. The character ( ) 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.

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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

枚举每个对面，再旋转比较剩下4个面，注意顺序。

完整代码：

#include <iostream>
using namespace std;

char str[15];

int judge(int a, int b, int c, int d, int e, int f)
{
	if (str[a] == str[0] && str[b] == str[5])
	{
		if (str[1] == str[c] && str[3] == str[d] && str[4] == str[e] && str[2] == str[f]) return true;
		if (str[1] == str[d] && str[3] == str[e] && str[4] == str[f] && str[2] == str[c]) return true;
		if (str[1] == str[e] && str[3] == str[f] && str[4] == str[c] && str[2] == str[d]) return true;
		if (str[1] == str[f] && str[3] == str[c] && str[4] == str[d] && str[2] == str[e]) return true;
		return false;
	}
	return false;
}

int main()
{
	while (cin >> str)
	{
		if (judge(6, 11, 7, 9, 10, 8)
		||judge(11, 6, 10, 9, 7, 8)
		||judge(7, 10, 11, 9, 6, 8)
		||judge(10, 7, 6, 9, 11, 8)
		||judge(8, 9, 11, 7, 6, 10)
		||judge(9, 8, 6, 7, 11, 10)) 
            cout << "TRUE" << endl;
		else cout << "FALSE" << endl;
	}
}