2014西安網路賽1006||hdu5012 bfs

http://acm.hdu.edu.cn/showproblem.php?pid=5012

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 a_i ≠ a_j and b_i ≠ b_j 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, a_i ≠ b_i). Ddy wants to make the two dices look the same from all directions(which means for all i, a_i = b_i) 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.

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <math.h>
#include <queue>
#include <stack>
#include <vector>
#include <algorithm>
using namespace std;

typedef long long LL;
const int INF=0x3f3f3f3f;
const int N=10005;

bool v[7][7][7][7][7][7];
struct note
{
    int a1,a2,a3,a4,a5,a6;
}z;

note L(note x)
{
    note y;
    y.a2=x.a5;
    y.a6=x.a2;
    y.a1=x.a6;
    y.a5=x.a1;
    y.a3=x.a3;
    y.a4=x.a4;
    return y;
}
note R(note x)
{
    note y;
    y.a5=x.a2;
    y.a2=x.a6;
    y.a6=x.a1;
    y.a1=x.a5;
    y.a3=x.a3;
    y.a4=x.a4;
    return y;
}
note F(note x)
{
    note y;
    y.a2=x.a3;
    y.a4=x.a2;
    y.a1=x.a4;
    y.a3=x.a1;
    y.a5=x.a5;
    y.a6=x.a6;
    return y;
}
note B(note x)
{
    note y;
    y.a3=x.a2;
    y.a2=x.a4;
    y.a4=x.a1;
    y.a1=x.a3;
    y.a5=x.a5;
    y.a6=x.a6;
    return y;
}
bool judge(note x)
{
    //printf("5555\n");
   // printf("%d %d %d %d %d %d\n",x.a1,x.a2,x.a3,x.a4,x.a5,x.a6);
    if(v[x.a1][x.a2][x.a3][x.a4][x.a5][x.a6]==0)
        return true;
    return false;
}
int bfs(note a)
{
    queue< pair<note,int> > q;
    q.push(make_pair(a,0));
    while(!q.empty())
    {
        note x=q.front().first;
        int ans=q.front().second;
        q.pop();
        if(x.a1==z.a1&&x.a3==z.a3&&x.a5==z.a5&&x.a2==z.a2&&x.a4==z.a4&&x.a6==z.a6)
            return ans;
        note y=L(x);
        if(judge(y))
        {
            //printf("888\n");
             v[y.a1][y.a2][y.a3][y.a4][y.a5][y.a6]=1;
             q.push(make_pair(y,ans+1));
        }
        y=R(x);
        if(judge(y))
        {
             v[y.a1][y.a2][y.a3][y.a4][y.a5][y.a6]=1;
             q.push(make_pair(y,ans+1));
        }
        y=F(x);
        if(judge(y))
        {
             v[y.a1][y.a2][y.a3][y.a4][y.a5][y.a6]=1;
             q.push(make_pair(y,ans+1));
        }
        y=B(x);
        if(judge(y))
        {
             v[y.a1][y.a2][y.a3][y.a4][y.a5][y.a6]=1;
             q.push(make_pair(y,ans+1));
        }
    }
    return -1;
}
int main()
{
    note x;
    while(~scanf("%d",&x.a1))
    {
        scanf("%d%d%d%d%d",&x.a2,&x.a3,&x.a4,&x.a5,&x.a6);
        scanf("%d%d%d%d%d%d",&z.a1,&z.a2,&z.a3,&z.a4,&z.a5,&z.a6);
        memset(v,false,sizeof(v));
        printf("%d\n",bfs(x));
    }
    return 0;
}