UVa

模拟

题意

给骰子涂色，用rbg表示颜色，骰子可以通过旋转变成一致

[ UVA - 253 (VJ) ]

思路

一开始思路很卡,枚举有误,借鉴了大神的思路才过

模拟骰子,固定两个面为底面不动(6种情况),剩下四个面旋转(4种情况),即可模拟枚举出4*6=24种情况

AC代码

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define maxn 20
char s[maxn],s1[maxn],s2[maxn];

int dr[6][6]={ {0,1,2,3,4,5},{5,4,2,3,1,0},{1,5,2,3,0,4},{4,0,2,3,5,1},{2,1,5,0,4,3},{3,1,0,5,4,2} };  //固定两个底的6种情况

void spin( char s2[], int a, int b, int c, int d ) //旋转 (*4)
{
    char t;
    t = s2[a];
    s2[a] = s2[b];
    s2[b] = s2[c];
    s2[c] = s2[d];
    s2[d] = t;
}

int judge()
{
    char temp[maxn] = {0};
    for(int i = 0; i < 6; i++)
    {
        for(int j = 0; j < 6; j++)
            temp[j] = s1[dr[i][j]];
        for( int j = 0; j < 4; j++ ){
            spin( temp, 1, 2, 4, 3 );
            if( strcmp(temp,s2) == 0 )
                return 1;
        }
    }
    return 0;
}

int main()
{
    while( scanf("%s",s) != EOF )
    {
        for( int i = 0 ; i < 6 ; i++ )
            s1[i] = s[i];
        for( int i = 0 ; i < 6 ; i++ )
            s2[i] = s[i+6];
        if( judge() )   puts("TRUE");
        else    puts("FALSE");
    }
    return 0;
}