hdu5012 Dice

Dice

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1931    Accepted Submission(s): 960


Problem Description

There are 2 special dices on the table. On each face of the dice, a distinct number was written. Consider a1.a2,a3,a4,a5,a6 to be numbers written on top face, bottom face, left face, right face,
front face and back face of dice A. Similarly, consider b1.b2,b3,b4,b5,b6 to be numbers on specific faces of dice B. It’s guaranteed that all numbers written on dices are integers no smaller
than 1 and no more than 6 while ai ≠ aj and bi ≠ bj for all i ≠ j. Specially, sum of numbers on opposite faces may not be 7.

At the beginning, the two dices may face different(which means there exist some i, ai ≠ bi). Ddy wants to make the two dices look the same from all directions(which means for all i, ai = bi) only by the following
four rotation operations.(Please read the picture for more information)



Now Ddy wants to calculate the minimal steps that he has to take to achieve his goal.

 

Input

There are multiple test cases. Please process till EOF. 

For each case, the first line consists of six integers a1,a2,a3,a4,a5,a6, representing the numbers on dice A. 

The second line consists of six integers b1,b2,b3,b4,b5,b6, representing the numbers on dice B.

 

Output

For each test case, print a line with a number representing the answer. If there’s no way to make two dices exactly the same, output -1.

 

Sample Input

1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 5 6 4 3
1 2 3 4 5 6
1 4 2 5 3 6

 

Sample Output

0
3
-1

 

Source

2014 ACM/ICPC Asia Regional Xi'an Online

只有四个方向，技巧：开一个六维的数组，对骰子的状态进行标记，如果已经进入队列了就不再push了；

                                   一直等到队列为空还没有找到与之相配的状态则ans=-1；

代码：

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
#define inf 0x3f3f3f3f

int v[7][7][7][7][7][7];

struct info
{
int a1,a2,a3,a4,a5,a6,num;
} x,y;

int ans=0;

queue<info>que;
void bfs()
{
while(!que.empty())  que.pop();
if(y.a1==x.a1&&y.a2==x.a2&&y.a3==x.a3&&y.a4==x.a4&&y.a5==x.a5&&y.a6==x.a6)

return ;

x.num=0;
que.push(x);

info B;
while(!que.empty())
{
info A=que.front();
que.pop();

B.num=A.num+1;
//左转
B.a1=A.a4,B.a2=A.a3,B.a3=A.a1,B.a4=A.a2,B.a5=A.a5,B.a6=A.a6;
if(B.a1==y.a1&&B.a2==y.a2&&B.a3==y.a3&&B.a4==y.a4&&B.a5==y.a5&&B.a6==y.a6)
{
ans=B.num;
return ;
}
else
{
if(!v[B.a1][B.a2][B.a3][B.a4][B.a5][B.a6])
{
que.push(B);
v[B.a1][B.a2][B.a3][B.a4][B.a5][B.a6]=1;
}
}

//右转
B.a1=A.a3,B.a2=A.a4,B.a3=A.a2,B.a4=A.a1,B.a5=A.a5,B.a6=A.a6;
if(B.a1==y.a1&&B.a2==y.a2&&B.a3==y.a3&&B.a4==y.a4&&B.a5==y.a5&&B.a6==y.a6)
{
ans=B.num;
return ;
}
else
{
if(!v[B.a1][B.a2][B.a3][B.a4][B.a5][B.a6])
{

que.push(B);
v[B.a1][B.a2][B.a3][B.a4][B.a5][B.a6]=1;
}
}
//上翻
B.a1=A.a5,B.a2=A.a6,B.a3=A.a3,B.a4=A.a4,B.a5=A.a2,B.a6=A.a1;
if(B.a1==y.a1&&B.a2==y.a2&&B.a3==y.a3&&B.a4==y.a4&&B.a5==y.a5&&B.a6==y.a6)
{
ans=B.num;
return ;
}
else
{
if(!v[B.a1][B.a2][B.a3][B.a4][B.a5][B.a6])
{
que.push(B);
v[B.a1][B.a2][B.a3][B.a4][B.a5][B.a6]=1;
}
}
//下翻
B.a1=A.a6,B.a2=A.a5,B.a3=A.a3,B.a4=A.a4,B.a5=A.a1,B.a6=A.a2;
if(B.a1==y.a1&&B.a2==y.a2&&B.a3==y.a3&&B.a4==y.a4&&B.a5==y.a5&&B.a6==y.a6)
{
ans=B.num;
return ;
}
else
{
if(!v[B.a1][B.a2][B.a3][B.a4][B.a5][B.a6])
{

que.push(B);
v[B.a1][B.a2][B.a3][B.a4][B.a5][B.a6]=1;
}
}
}
ans=-1;
return ;

}

int main()
{
while(~scanf("%d%d%d%d%d%d",&x.a1,&x.a2,&x.a3,&x.a4,&x.a5,&x.a6))
{
scanf("%d%d%d%d%d%d",&y.a1,&y.a2,&y.a3,&y.a4,&y.a5,&y.a6);
memset(v,0,sizeof(v));
ans=0;
bfs();
printf("%d\n",ans);
}
return 0;
}