水题：UVa253-Cube painting

Cube painting

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. Since a cube has 6 faces, our machine can paint a face-numbered cube in 36 = 729 different ways. When ignoring the face-numbers, the number of different paintings is much less, because a cube can be rotated. See example below. Wedenoteapaintedcubebyastringof6characters, whereeach character is a ‘b’, ‘r’, or ‘g’. The i-th character (1 ≤ i ≤ 6) 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.

Input

The input of your program is a text file 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

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解题心得：

1. 很简单的一个题，只要比较三个对面是否相等就可以了，只要三个对面相等，怎么安排都是一样的六面体。

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反正都这么简单，代码就瞎写的

#include<bits/stdc++.h>
using namespace std;
const int maxn = 20;
char ch[maxn];
int main()
{
    while(scanf("%s",ch+1) != EOF)
    {
        bool flag = false;
        for(int i=1;i<=6;i++)
        {
            int k;
            if(i == 1)
                k = 6;
            else if(i == 2)
                k = 5;
            else if(i == 3)
                k = 4;
            else if(i == 4)
                k = 3;
            else if(i == 5)
                k = 2;
            else if(i == 6)
                k = 1;
            if(ch[i] == 'A')
                continue;
            char now1,now2;
            now1 = ch[i];
            now2 = ch[k];
            ch[i] = 'A',ch[k] = 'A';
            flag = false;
            for(int j=7;j<=12;j++)
            {
                if(ch[j] == now1)
                {
                    if(j == 7 && now2 == ch[12])
                    {
                        ch[7] = ch[12] = 'A';
                        flag = true;
                    }
                    else if(j == 8 && now2 == ch[11])
                    {
                        flag = true;
                        ch[8] = ch[11] = 'A';
                    }
                    else if(j == 9 && now2 == ch[10])
                    {
                        flag = true;
                        ch[9] = ch[10] = 'A';
                    }
                    else if(j == 10 && now2 == ch[9])
                    {
                        flag = true;
                        ch[10] = ch[9] = 'A';
                    }
                    else if(j == 11 && now2 == ch[8])
                    {
                        flag = true;
                        ch[8] = ch[11] = 'A';
                    }
                    else if(j == 12 && now2 == ch[7])
                    {
                        flag = true;
                        ch[7] = ch[12] = 'A';
                    }
                }
                if(flag)
                    break;
            }
            if(!flag)
            {
                printf("FALSE\n");
                break;
            }
        }
        if(!flag)
            continue;
        else
            printf("TRUE\n");
    }
    return 0;
}