拧魔方问题-模拟法

问题描述：魔方是一种常见的玩具。2010年7月，美国加利福尼亚州科学家利用计算机证明任意组合的魔方均可以在20步之内还原。作为一个入门级的程序员，我们决定先写一个验证魔方是否复原的程序。对于魔方的一个操作，我们用一个字母来表示。将魔方的一个面正对玩家，就有了前后上下左右六个面，分别用F(Front)，B(Back)，U(Up)，D(Down):，L(Left)，R(Right)来表示将这个面顺时针旋转90度，具体玩魔方的时候将右手覆盖到对应的面上，这六个操作时右手的旋转方向都是相同的。同时用X，Y，Z，表示顺时针旋转中间一层，分别对应U，R，F。具体情况可以参照下图。与这九个操作对应的还有f,b,u,d,l,r,x,y,z，表示逆时针旋转。

现在我们给出一个操作序列，问在这么旋转之后，魔方是否和原来的时候完全一样。比如UXd被认为是不一样。输入为一个长度不超过200的字符串，仅包含之上定义的18个字母。如果能复原，输出Yes，否则输出No。

测试用例输入：RUur
测试用例输出：Yes

#include<stdio.h>

char Cube[6][9];

void Init  ( void );
void Front ( void );
void Back  ( void );
void Up    ( void );
void Down  ( void );
void Left  ( void );
void Right ( void );
void X     ( void );
void Y     ( void );
void Z     ( void );

int main( )
{
	int  i , j ;
	char ch;

	Init( );

	while( (ch=getchar())!='\n' )
	{
		if( ch=='F' )
			Front( );

		else if( ch=='B' )
			Back( );
	
		else if( ch=='U' )
			Up( );

		else if( ch=='D' )
			Down( );

		else if( ch=='L' )
			Left( );

		else if( ch=='R' )
			Right( );

		else if( ch=='X' )
			X( );

		else if( ch=='Y' )
			Y( );

		else if( ch=='Z' )
			Z( );

		else if( ch=='f' )
		{
			for( i=1 ; i<=3 ; i++ )
				Front( );
		}

		else if( ch=='b' )
		{
			for( i=1 ; i<=3 ; i++ )
			Back( );
		}

		else if( ch=='u' )
		{
			for( i=1 ; i<=3 ; i++ )
				Up( );
		}

		else if( ch=='d' )
		{
			for( i=1 ; i<=3 ; i++ )
				Down( );
		}

		else if( ch=='l' )
		{
			for( i=1 ; i<=3 ; i++ )
				Left( );
		}

		else if( ch=='r' )
		{
			for( i=1 ; i<=3 ; i++ )
				Right( );
		}

		else if( ch=='x' )
		{
			for( i=1 ; i<=3 ; i++ )
				X( );
		}

		else if( ch=='y' )
		{
			for( i=1 ; i<=3 ; i++ )
				Y( );
		}

		else if( ch=='z' )
		{
			for( i=1 ; i<=3 ; i++ )
				Z( );
		}
	}

	for( i=0 ; i<6 ; i++ )
	{
		for( j=0 ; j<9 ; j++ )
			if( Cube[i][j] != Cube[i][0] )
			{
				printf("No\n");

				return 0;
			}
	}

	printf("Yes\n");
	
	return 0;
}


void Init( void )
{
	int i , j , k=1;

	for( i=0 , k=1 ; i<6 ; i++ , k++ )
	{
		for( j=0 ; j<9 ; j++ )
			Cube[i][j] = k;
	}
}


void Front( void )
{
	int k , j , Color[12];

	for( k=0 ; k<9 ; k++ )
		Color[k] = Cube[0][k];

	Cube[0][0] = Color[6];
	Cube[0][1] = Color[3];
	Cube[0][2] = Color[0];

	Cube[0][3] = Color[7];
	Cube[0][4] = Color[4];
	Cube[0][5] = Color[1];

	Cube[0][6] = Color[8];
	Cube[0][7] = Color[5];
	Cube[0][8] = Color[2];

	k=0;

	for( j=6 ; j<9 ; j++ )
		Color[k++] = Cube[4][j];

	for( j=0  ; j<9 ; j+=3 )
		Color[k++] = Cube[3][j];

	for( j=8 ; j>0 ; j-=3 )
		Color[k++] = Cube[5][j];

	for( j=8 ; j>0 ; j-=3 )
		Color[k++] = Cube[2][j];

	k=0;

	Cube[3][0] = Color[k++];
	Cube[3][3] = Color[k++];
	Cube[3][6] = Color[k++];

	Cube[5][8] = Color[k++];
	Cube[5][5] = Color[k++];
	Cube[5][2] = Color[k++];

	Cube[2][8] = Color[k++];
	Cube[2][5] = Color[k++];
	Cube[2][2] = Color[k++];

	Cube[4][6] = Color[k++];
	Cube[4][7] = Color[k++];
	Cube[4][8] = Color[k++];
}



void Back( void )
{
	int k , j , Color[12];

	for( k=0 ; k<9 ; k++ )
		Color[k] = Cube[1][k];

	Cube[1][0] = Color[6];
	Cube[1][1] = Color[3];
	Cube[1][2] = Color[0];

	Cube[1][3] = Color[7];
	Cube[1][4] = Color[4];
	Cube[1][5] = Color[1];

	Cube[1][6] = Color[8];
	Cube[1][7] = Color[5];
	Cube[1][8] = Color[2];

	k=0;

	for( j=2 ; j>=0 ; j-- )
		Color[k++] = Cube[4][j];

	for( j=0  ; j<9 ; j+=3 )
		Color[k++] = Cube[2][j];

	for( j=0  ; j<9 ; j+=3 )
		Color[k++] = Cube[5][j];

	for( j=8 ; j>0 ; j-=3 )
		Color[k++] = Cube[3][j];

	k=0;

	Cube[2][0] = Color[k++];
	Cube[2][3] = Color[k++];
	Cube[2][6] = Color[k++];

	Cube[5][0] = Color[k++];
	Cube[5][3] = Color[k++];
	Cube[5][6] = Color[k++];

	Cube[3][8] = Color[k++];
	Cube[3][5] = Color[k++];
	Cube[3][2] = Color[k++];

	Cube[4][2] = Color[k++];
	Cube[4][1] = Color[k++];
	Cube[4][0] = Color[k++];
}


void Up( void )
{
	int k , j , Color[12];

	for( k=0 ; k<9 ; k++ )
		Color[k] = Cube[4][k];

	Cube[4][0] = Color[6];
	Cube[4][1] = Color[3];
	Cube[4][2] = Color[0];

	Cube[4][3] = Color[7];
	Cube[4][4] = Color[4];
	Cube[4][5] = Color[1];

	Cube[4][6] = Color[8];
	Cube[4][7] = Color[5];
	Cube[4][8] = Color[2];

	k=0;

	for( j=2 ; j>=0 ; j-- )
		Color[k++] = Cube[0][j];

	for( j=2  ; j>=0 ; j-- )
		Color[k++] = Cube[2][j];

	for( j=2  ; j>=0 ; j-- )
		Color[k++] = Cube[1][j];

	for( j=2  ; j>=0 ; j-- )
		Color[k++] = Cube[3][j];

	k=0;

	Cube[2][2] = Color[k++];
	Cube[2][1] = Color[k++];
	Cube[2][0] = Color[k++];

	Cube[1][2] = Color[k++];
	Cube[1][1] = Color[k++];
	Cube[1][0] = Color[k++];

	Cube[3][2] = Color[k++];
	Cube[3][1] = Color[k++];
	Cube[3][0] = Color[k++];

	Cube[0][2] = Color[k++];
	Cube[0][1] = Color[k++];
	Cube[0][0] = Color[k++];
}


void Down( void )
{
	int k , j , Color[12];

	for( k=0 ; k<9 ; k++ )
		Color[k] = Cube[5][k];

	Cube[5][0] = Color[6];
	Cube[5][1] = Color[3];
	Cube[5][2] = Color[0];

	Cube[5][3] = Color[7];
	Cube[5][4] = Color[4];
	Cube[5][5] = Color[1];

	Cube[5][6] = Color[8];
	Cube[5][7] = Color[5];
	Cube[5][8] = Color[2];

	k=0;

	for( j=6 ; j<=8 ; j++ )
		Color[k++] = Cube[0][j];

	for( j=6 ; j<=8 ; j++ )
		Color[k++] = Cube[3][j];

	for( j=6 ; j<=8 ; j++ )
		Color[k++] = Cube[1][j];

	for( j=6 ; j<=8 ; j++ )
		Color[k++] = Cube[2][j];

	k=0;

	Cube[3][6] = Color[k++];
	Cube[3][7] = Color[k++];
	Cube[3][8] = Color[k++];

	Cube[1][6] = Color[k++];
	Cube[1][7] = Color[k++];
	Cube[1][8] = Color[k++];

	Cube[2][6] = Color[k++];
	Cube[2][7] = Color[k++];
	Cube[2][8] = Color[k++];

	Cube[0][6] = Color[k++];
	Cube[0][7] = Color[k++];
	Cube[0][8] = Color[k++];
}


void Left( void )
{
	int k , j , Color[12];

	for( k=0 ; k<9 ; k++ )
		Color[k] = Cube[2][k];

	Cube[2][0] = Color[6];
	Cube[2][1] = Color[3];
	Cube[2][2] = Color[0];

	Cube[2][3] = Color[7];
	Cube[2][4] = Color[4];
	Cube[2][5] = Color[1];

	Cube[2][6] = Color[8];
	Cube[2][7] = Color[5];
	Cube[2][8] = Color[2];

	k=0;

	for( j=0 ; j<9 ; j+=3 )
		Color[k++] = Cube[4][j];

	for( j=0 ; j<9 ; j+=3 )
		Color[k++] = Cube[0][j];

	for( j=2 ; j>=0 ; j-- )
		Color[k++] = Cube[5][j];

	for( j=8 ; j>=0 ; j-=3 )
		Color[k++] = Cube[1][j];

	k=0;

	Cube[0][0] = Color[k++];
	Cube[0][3] = Color[k++];
	Cube[0][6] = Color[k++];

	Cube[5][2] = Color[k++];
	Cube[5][1] = Color[k++];
	Cube[5][0] = Color[k++];

	Cube[1][8] = Color[k++];
	Cube[1][5] = Color[k++];
	Cube[1][2] = Color[k++];

	Cube[4][0] = Color[k++];
	Cube[4][3] = Color[k++];
	Cube[4][6] = Color[k++];
}


void Right( void )
{
	int k , j , Color[12];

	for( k=0 ; k<9 ; k++ )
		Color[k] = Cube[3][k];

	Cube[3][0] = Color[6];
	Cube[3][1] = Color[3];
	Cube[3][2] = Color[0];

	Cube[3][3] = Color[7];
	Cube[3][4] = Color[4];
	Cube[3][5] = Color[1];

	Cube[3][6] = Color[8];
	Cube[3][7] = Color[5];
	Cube[3][8] = Color[2];

	k=0;

	for( j=8 ; j>=0 ; j-=3 )
		Color[k++] = Cube[4][j];

	for( j=0 ; j<9 ; j+=3 )
		Color[k++] = Cube[1][j];

	for( j=6 ; j<9 ; j++ )
		Color[k++] = Cube[5][j];

	for( j=8 ; j>=0 ; j-=3 )
		Color[k++] = Cube[0][j];

	k=0;

	Cube[1][0] = Color[k++];
	Cube[1][3] = Color[k++];
	Cube[1][6] = Color[k++];

	Cube[5][6] = Color[k++];
	Cube[5][7] = Color[k++];
	Cube[5][8] = Color[k++];

	Cube[0][8] = Color[k++];
	Cube[0][5] = Color[k++];
	Cube[0][2] = Color[k++];

	Cube[4][8] = Color[k++];
	Cube[4][5] = Color[k++];
	Cube[4][2] = Color[k++];
}


void X( void )
{
	int k , j , Color[12];

	k=0;

	for( j=5 ; j>=3 ; j-- )
		Color[k++] = Cube[0][j];

	for( j=5 ; j>=3 ; j-- )
		Color[k++] = Cube[2][j];

	for( j=5 ; j>=3 ; j-- )
		Color[k++] = Cube[1][j];

	for( j=5 ; j>=3 ; j-- )
		Color[k++] = Cube[3][j];

	k=0;

	Cube[2][5] = Color[k++];
	Cube[2][4] = Color[k++];
	Cube[2][3] = Color[k++];

	Cube[1][5] = Color[k++];
	Cube[1][4] = Color[k++];
	Cube[1][3] = Color[k++];

	Cube[3][5] = Color[k++];
	Cube[3][4] = Color[k++];
	Cube[3][3] = Color[k++];

	Cube[0][5] = Color[k++];
	Cube[0][4] = Color[k++];
	Cube[0][3] = Color[k++];
}


void Y( void )
{
	int k , j , Color[12];

	k=0;

	for( j=7 ; j>=0 ; j-=3 )
		Color[k++] = Cube[4][j];

	for( j=1 ; j<9 ; j+=3 )
		Color[k++] = Cube[1][j];

	for( j=3 ; j<=5 ; j++ )
		Color[k++] = Cube[5][j];

	for( j=7 ; j>=0 ; j-=3 )
		Color[k++] = Cube[0][j];

	k=0;

	Cube[1][1] = Color[k++];
	Cube[1][4] = Color[k++];
	Cube[1][7] = Color[k++];

	Cube[5][3] = Color[k++];
	Cube[5][4] = Color[k++];
	Cube[5][5] = Color[k++];

	Cube[0][7] = Color[k++];
	Cube[0][4] = Color[k++];
	Cube[0][1] = Color[k++];

	Cube[4][7] = Color[k++];
	Cube[4][4] = Color[k++];
	Cube[4][1] = Color[k++];
}


void Z( void )
{
	int k , j , Color[12];

	k=0;

	for( j=3 ; j<=5 ; j++ )
		Color[k++] = Cube[4][j];

	for( j=1 ; j<9 ; j+=3 )
		Color[k++] = Cube[3][j];

	for( j=7 ; j>=0 ; j-=3 )
		Color[k++] = Cube[5][j];

	for( j=7 ; j>=0 ; j-=3 )
		Color[k++] = Cube[2][j];

	k=0;

	Cube[3][1] = Color[k++];
	Cube[3][4] = Color[k++];
	Cube[3][7] = Color[k++];

	Cube[5][7] = Color[k++];
	Cube[5][4] = Color[k++];
	Cube[5][1] = Color[k++];

	Cube[2][7] = Color[k++];
	Cube[2][4] = Color[k++];
	Cube[2][1] = Color[k++];

	Cube[4][3] = Color[k++];
	Cube[4][4] = Color[k++];
	Cube[4][5] = Color[k++];
}

代码分析：采用了模拟法的算法，代码结构流程非常清晰，思路简单，就是根据题目所示的意思来模拟要解决问题的步骤，一个简单的方法就是拿一个真正的魔方或者纸质的方体，并标上数字，这样比起纯空间想象要容易得多，错误率得以降低，不过代码的18的函数定义非常相似，可想方法来给予优化。请问大家这里有什么更好的算法或者可以优化这18个函数尽量简洁一点的方法么？