题意：

给出一个2X2X2的魔方，再给一个限定的步骤长度，不超过该长度最多能能使几个面拼成功。

纯粹模拟题，搞清楚几个面的变换关系，并化简步骤，三种旋转方式，两种旋转方向。bfs,dfs都可以。

#include <stdio.h>
#include <string.h>
#define maxn 300000
int twist[3][3][4]=
{
	{{1,7,17,21}, {3,13,19,23}, {9,8,14,15}},
	{{2,11,17,8}, {3,5,16,14},  {6,12,13,7}},
	{{0,2,3,1},   {23,4,6,8},   {22,5,7,9} },
};
struct P{
	int state[24];
	int step;
}p[maxn];
int usage[24];
int head,tail,N,ans;
int judge(int *a)
{
	int ans=0;
	if(a[0]==a[1] && a[0]==a[2]  && a[0]==a[3])
		ans++;
	if(a[4]==a[5] && a[4]==a[10] && a[4]==a[11])
		ans++;
	if(a[6]==a[7] && a[6]==a[12] && a[6]==a[13])
		ans++;
	if(a[8]==a[9] && a[8]==a[14] && a[8]==a[15])
		ans++;
	if(a[16]==a[17]&&a[16]==a[18]&& a[16]==a[19])
		ans++;
	if(a[20]==a[21]&&a[20]==a[22]&& a[20]==a[23])
		ans++;
	return ans;
}
int bfs()
{
	head=0,tail=1;
	int i,j,k;
	int index,tt;
	p[0].step=0;
	/*	for(k=0;k<24;k++)
		    printf("%2d ",k);
		printf("\n");
		for(k=0;k<24;k++)
		    printf("%2d ",p[0].state[k]);
		printf("\n");*/
	ans=judge(p[0].state);
	while(head<tail)
	{
		if(p[head].step==N||ans==6)
			break;
		
		for(i=0;i<3;i++)
		{
			memcpy(usage,p[head].state,sizeof(usage));
			for(j=0;j<3;j++)
			{
				for(k=1;k<4;k++)
				{
					index=twist[i][j][k];
					tt=usage[twist[i][j][0]];
					usage[twist[i][j][0]]=usage[index];
					usage[index]=tt;
				}
			}
			k=judge(usage);
			ans=ans>k?ans:k;
			if(p[head].step + 1 < N)
			{
				memcpy(p[tail].state,usage,sizeof(usage));
				p[tail++].step=p[head].step+1;
			}

			for(j=0;j<3;j++)
			{
				for(k=1;k<4;k++)
				{
					index=twist[i][j][k];
					tt=usage[twist[i][j][0]];
					usage[twist[i][j][0]]=usage[index];
					usage[index]=tt;
				}
			}
			for(j=0;j<3;j++)
			{
				for(k=1;k<4;k++)
				{
					index=twist[i][j][k];
					tt=usage[twist[i][j][0]];
					usage[twist[i][j][0]]=usage[index];
					usage[index]=tt;
				}
			}
			k=judge(usage);
			ans=ans>k?ans:k;
			if(p[head].step + 1 < N)
			{
				memcpy(p[tail].state,usage,sizeof(usage));
				p[tail++].step=p[head].step+1;
			}
		}
		head++;
	}
			
	
	return ans;
}
int main(){
	//freopen("D:\\a.txt","w",stdout);
   int i,n;
   while(scanf("%d",&N)!=EOF)
   {
	   for(i=0;i<24;i++)
		   scanf("%d",p[0].state+i);
	   n=bfs();
	   printf("%d\n",n);
   }

    return 0;
}

2013 长沙现场赛 K (Pocket Cube)