1 条题解
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仔细想一想，魔方一共有6个面，转每个面可以看成是转一个面，首先就要把每个面表示出来，用指针，在函数中可以改每个面的数。 分别把每个面当成正面，上下左右重新弄一下，然后进入顺时针或者逆时针的函数。由于指定一个面是正面，其他的面不是简单的平移关系，这时候就要把正面周围的面找出来，luckily，我们只要找和正面周围的一行或一列就行，所以找其他的面时，只要改变与正面有关系的地方就行了完事之后就变回来
#include<iostream> #include<cstdio> using namespace std; char sa[10][10],sb[10][10],sc[10][10]; void change0( char (*to)[5], char (*bo)[5] ) { for (int i = 1; i <= 3; i++) for (int j = 1; j <= 3; j++) { sa[i][j] = to[i][j]; sb[i][j] = bo[i][j]; } for (int i = 1; i <= 3; i++) { to[3][i] = sa[i][1]; bo[1][i] = sb[4-i][1]; } return; } void dischange0( char (*to)[5], char (*bo)[5] ) { for (int i = 1; i <= 3; i++) { sa[i][1] = to[3][i]; sb[i][1] = bo[1][4-i]; } for (int i = 1; i <= 3; i++) for (int j = 1; j <= 3; j++) { to[i][j] = sa[i][j]; bo[i][j] = sb[i][j]; } return; } void change2( char (*to)[5], char (*bo)[5] ) { for (int i = 1; i <= 3; i++) for (int j = 1; j <= 3; j++) { sa[i][j] = to[i][j]; sb[i][j] = bo[i][j]; } for (int i = 1; i <= 3; i++) { to[3][i] = sa[4-i][3]; bo[1][i] = sb[i][3]; } return; } void dischange2( char (*to)[5], char (*bo)[5] ) { for (int i = 1; i <= 3; i++) { sa[i][3] = to[3][4-i]; sb[i][3] = bo[1][i]; } for (int i = 1; i <= 3; i++) for (int j = 1; j <= 3; j++) { to[i][j] = sa[i][j]; bo[i][j] = sb[i][j]; } return; } void change3( char (*to)[5], char (*bo)[5] ) { for (int i = 1; i <= 3; i++) for (int j = 1; j <= 3; j++) { sa[i][j] = to[i][j]; sb[i][j] = bo[i][j]; } for (int i = 1; i <= 3; i++) { to[3][i] = sa[1][4-i]; bo[1][i] = sb[3][4-i]; } return; } void dischange3( char (*to)[5], char (*bo)[5] ) { for (int i = 1; i <= 3; i++) { sa[1][i] = to[3][4-i]; sb[3][i] = bo[1][4-i]; } for (int i = 1; i <= 3; i++) for (int j = 1; j <= 3; j++) { to[i][j] = sa[i][j]; bo[i][j] = sb[i][j]; } return; } void change4(char (*le)[5], char (*ri)[5], char (*ba)[5]) { for (int i = 1; i <= 3; i++) for (int j = 1; j <= 3; j++) { sa[i][j] = le[i][j]; sb[i][j] = ri[i][j]; sc[i][j] = ba[i][j]; } for (int i = 1; i <= 3; i++) { le[i][3] = sa[1][i]; ri[i][1] = sb[1][4-i]; ba[3][i] = sc[1][4-i]; } return; } void dischange4(char (*le)[5], char (*ri)[5], char (*ba)[5]) { for (int i = 1; i <= 3; i++) { sa[1][i] = le[i][3]; sb[1][i] = ri[4-i][1]; sc[1][i] = ba[3][4-i]; } for (int i = 1; i <= 3; i++) for (int j = 1; j <= 3; j++) { le[i][j] = sa[i][j]; ba[i][j] = sc[i][j]; ri[i][j] = sb[i][j]; } return; } void change5(char (*le)[5], char (*ri)[5], char (*ba)[5]) { for (int i = 1; i <= 3; i++) for (int j = 1; j <= 3; j++) { sa[i][j] = le[i][j]; sb[i][j] = ri[i][j]; sc[i][j] = ba[i][j]; } for (int i = 1; i <= 3; i++) { le[i][3] = sa[3][4-i]; ri[i][1] = sb[3][i]; ba[1][i] = sc[3][4-i]; } return; } void dischange5(char (*le)[5], char (*ri)[5], char (*ba)[5]) { for (int i = 1; i <= 3; i++) { sa[3][i] = le[4-i][3]; sb[3][i] = ri[i][1]; sc[3][i] = ba[1][4-i]; } for (int i = 1; i <= 3; i++) for (int j = 1; j <= 3; j++) { le[i][j] = sa[i][j]; ri[i][j] = sb[i][j]; ba[i][j] = sc[i][j]; } return; } void clockwise(char (*fr)[5],char (*le)[5], char (*ri)[5], char (*to)[5], char (*bo)[5]) { char s[10][10]; for (int i = 1; i <= 3; i++) for (int j = 1; j <= 3; j++) s[i][j] = fr[i][j]; for (int i = 1; i <= 3; i++) for (int j = 1; j <= 3; j++) fr[j][4-i] = s[i][j]; char x,y,z; x = le[1][3]; y = le[2][3]; z = le[3][3]; for (int i = 1; i <= 3; i++) le[i][3] = bo[1][i]; for (int i = 1; i <= 3; i++) bo[1][i] = ri[4-i][1]; for (int i = 1; i <= 3; i++) ri[i][1] = to[3][i]; to[3][1] = z; to[3][2] = y; to[3][3] = x; return; } void counterclockwise(char (*fr)[5],char (*le)[5], char (*ri)[5], char (*to)[5], char (*bo)[5]) { char s[10][10]; for (int i = 1; i <= 3; i++) for (int j = 1; j <= 3; j++) s[i][j] = fr[i][j]; for (int i = 1; i <= 3; i++) for (int j = 1; j <= 3; j++) fr[4-j][i] = s[i][j]; char x,y,z; x = le[1][3]; y = le[2][3]; z = le[3][3]; for (int i = 1; i <= 3; i++) le[i][3] = to[3][4-i]; for (int i = 1; i <= 3; i++) to[3][i] = ri[i][1]; for (int i = 1; i <= 3; i++) ri[i][1] = bo[1][4-i]; bo[1][1] = x; bo[1][2] = y; bo[1][3] = z; return; } int main() { char fr[10][5],ba[10][5],le[10][5],ri[10][5],to[10][5],bo[10][5]; char map[100][100]; int n,m,t,k,x,y; //freopen("yu.in","r",stdin); //freopen("yu.out","w",stdout); scanf("%d",&t); k = 0; while(t--) { for (int i = 0; i <= 9; i++) cin.getline(map[i],100); for (int i = 1; i <= 3; i++) for (int j = 1; j <= 3; j++) { to[i][j] = map[i][2*j+4]; le[i][j] = map[3+i][2*j-2]; fr[i][j] = map[3+i][4+j*2]; ri[i][j] = map[3+i][10+2*j]; ba[i][j] = map[3+i][16+2*j]; bo[i][j] = map[6+i][4+j*2]; } scanf("%d",&n); for (int i = 1; i <= n; i++) { scanf("%d%d",&x,&y); switch (x) { case 0 : change0(to,bo); if (y == 1) clockwise(le,ba,fr,to,bo); else counterclockwise(le,ba,fr,to,bo); dischange0(to,bo); break; case 1 : if (y == 1) clockwise(fr,le,ri,to,bo); else counterclockwise(fr,le,ri,to,bo); break; case 2 : change2(to,bo);if (y == 1) clockwise(ri,fr,ba,to,bo); else counterclockwise(ri,fr,ba,to,bo); dischange2(to,bo); break; case 3 : change3(to,bo);if (y == 1) clockwise(ba,ri,le,to,bo); else counterclockwise(ba,ri,le,to,bo); dischange3(to,bo); break; case 4 : change4(le,ri,ba);if (y == 1) clockwise(to,le,ri,ba,fr); else counterclockwise(to,le,ri,ba,fr); dischange4(le,ri,ba); break; case 5 : change5(le,ri,ba);if (y == 1) clockwise(bo,le,ri,fr,ba); else counterclockwise(bo,le,ri,fr,ba); dischange5(le,ri,ba); break; } } k++; printf("Scenario #%d:\n",k); for (int i = 1; i <= 3; i++) printf(" %c %c %c\n",to[i][1],to[i][2],to[i][3]); for (int i = 1; i <= 3; i++) printf("%c %c %c %c %c %c %c %c %c %c %c %c\n",le[i][1],le[i][2],le[i][3],fr[i][1],fr[i][2],fr[i][3],ri[i][1],ri[i][2],ri[i][3],ba[i][1],ba[i][2],ba[i][3]); for (int i = 1; i <= 3; i++) printf(" %c %c %c\n",bo[i][1],bo[i][2],bo[i][3]); printf("\n"); } return 0; }
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