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  • UVA-253-Cube painting

    Description

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    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.

    picture21

    Figure 1.

    Since a cube has 6 faces, our machine can paint a face-numbered cube in tex2html_wrap_inline126 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 br, or g. The tex2html_wrap_inline128 character ( tex2html_wrap_inline130 ) 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 tex2html_wrap_inline134 , the one changes into the other.

    tex2html_wrap138tex2html_wrap140

    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


    #include <cstdio>
    #include <cstring>
    #include <iostream>
    #include <algorithm>
    #include <queue>
    #include <string>
    using namespace std;
    //全面检索一次如果该还不能让flog变为true 那么这2个cube不相同;
    int main()
    {
        char a[15];
        while(scanf("%s",a)!=-1)
        {
            bool flog;
            for (int i=0; i < 3; i++)
            {
                flog=false;
                for (int j=0; j < 3; j++)
                {
                    if(a[i]==a[j+6]&&a[5-i]==a[11-j])
                    {
                        a[j+6]='a',a[11-j]='a';
                        flog=true;
                        break;
                    }
                    else if(a[i]==a[11-j]&&a[5-i]==a[j+6])
                    {
                        a[j+6]='a',a[11-j]='a';
                        flog=true;
                        break;
                    }
                }
                if(!flog) break;
            }
            if(flog) printf("TRUE
    ");
            else printf("FALSE
    ");
        }
        return 0;
    }









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  • 原文地址:https://www.cnblogs.com/chinashenkai/p/9451409.html
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