hdu 5012 Dice（bfs）

Dice

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

Total Submission(s): 577 Accepted Submission(s): 335



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

用队列进行广搜就行了，开一个6维数组。

#include"stdio.h"
#include"string.h"
#include"queue"
#include"iostream"
#include"algorithm"
using namespace std;
#define N  7
#define LL __int64
int p

;
int a
,b
;
struct node
{
int c
,t;
friend bool operator<(node a,node b)
{
return a.t>b.t;
}
};
int bfs()
{
int i,flag=-1;
memset(p,0,sizeof(p));
priority_queue<node>q;
node cur,next;
for(i=1;i<N;i++)
cur.c[i]=a[i];
cur.t=0;
q.push(cur);
p[a[1]][a[2]][a[3]][a[4]][a[5]][a[6]]=1;
while(!q.empty())
{
cur=q.top();
q.pop();
for(i=1;i<N;i++)
if(b[i]!=cur.c[i])
break;
if(i==N)
{
flag=cur.t;
break;
}
next.t=cur.t+1;
next.c[1]=cur.c[4];
next.c[2]=cur.c[3];
next.c[3]=cur.c[1];
next.c[4]=cur.c[2];
next.c[5]=cur.c[5];
next.c[6]=cur.c[6];
for(i=1;i<N;i++)
a[i]=next.c[i];
if(p[a[1]][a[2]][a[3]][a[4]][a[5]][a[6]]==0)
{
q.push(next);
p[a[1]][a[2]][a[3]][a[4]][a[5]][a[6]]=1;
}

next.c[1]=cur.c[3];
next.c[2]=cur.c[4];
next.c[3]=cur.c[2];
next.c[4]=cur.c[1];
for(i=1;i<N;i++)
a[i]=next.c[i];
if(p[a[1]][a[2]][a[3]][a[4]][a[5]][a[6]]==0)
{
q.push(next);
p[a[1]][a[2]][a[3]][a[4]][a[5]][a[6]]=1;
}

next.c[1]=cur.c[6];
next.c[2]=cur.c[5];
next.c[3]=cur.c[3];
next.c[4]=cur.c[4];
next.c[5]=cur.c[1];
next.c[6]=cur.c[2];
for(i=1;i<N;i++)
a[i]=next.c[i];
if(p[a[1]][a[2]][a[3]][a[4]][a[5]][a[6]]==0)
{
q.push(next);
p[a[1]][a[2]][a[3]][a[4]][a[5]][a[6]]=1;
}

next.c[1]=cur.c[5];
next.c[2]=cur.c[6];
next.c[5]=cur.c[2];
next.c[6]=cur.c[1];
for(i=1;i<N;i++)
a[i]=next.c[i];
if(p[a[1]][a[2]][a[3]][a[4]][a[5]][a[6]]==0)
{
q.push(next);
p[a[1]][a[2]][a[3]][a[4]][a[5]][a[6]]=1;
}
}
return flag;
}
int main()
{
int i;
while(scanf("%d%d%d%d%d%d",&a[1],&a[2],&a[3],&a[4],&a[5],&a[6])!=-1)
{
for(i=1;i<N;i++)
scanf("%d",&b[i]);
int ans=bfs();
printf("%d\n",ans);
}
return 0;
}