算法竞赛入门经典(第二版)-刘汝佳-第四章函数与递归骰子上色Uva253

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本文介绍了一种算法，用于判断两个六面体通过旋转是否能互相匹配颜色布局。输入为两组六面体的颜色字符串表示，输出为布尔值表明是否匹配。通过排序和比较相对面的颜色来简化匹配过程。

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, or g. 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

只要每个面以及它的相对的面两个色子都一样即可

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int c1(const void *a,const void *b)
{
    return strcmp((char *)a,(char *)b);
}

int main()
{
    char s[13],s1[7]={},s2[7]={},cmp1[3][3]={},cmp2[3][3]={};
    int i,j,k,sum=0;
    while(gets(s))
    {
    for(i=0;i<6;i++)
    {
        s1[i]=s[i];
    }
    for(i=6;i<12;i++)
    {
        s2[i-6]=s[i];
    }
    for(k=0,i=0,j=5;i<3;i++,j--,k++)
    {
        cmp1[k][0]=s1[i];
        cmp1[k][1]=s1[j];
    }
    for(k=0,i=0,j=5;i<3;i++,j--,k++)
    {
        cmp2[k][0]=s2[i];
        cmp2[k][1]=s2[j];
    }
    for(i=0;i<3;i++)
    {
        qsort(cmp1[i],2,sizeof(char),c1);
        qsort(cmp2[i],2,sizeof(char),c1);
    }

    for(i=0;i<3;i++)
    {
        for(j=0;j<3;j++)
        {
            if(strcmp(cmp2[i],cmp1[j])==0&&strcmp(cmp1[j],"00")!=0)
            {
                strcpy(cmp2[i],"00");
                strcpy(cmp1[j],"00");
                sum++;
            }

        }
    }


    if(sum==3)
        printf("TRUE\n");
    else
        printf("FALSE\n");
        sum=0;
    }
}