题目链接：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.

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

0
3
-1

先贴一发队友写的。等会再补上本弱的！

代码如下：

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

struct saizi
{

    int b1,b2,b3,b4,b5,b6;
    int size;
    saizi() {}
    saizi(int a1,int a2,int a3,int a4,int a5,int a6,int si):b1(a1),b2(a2),b3(a3),b4(a4),b5(a5),b6(a6),size(si) {};

    bool operator == (const saizi& b) const
    {
        if(b.b1==b1&&b.b2==b2&&b.b3==b3&&b.b4==b4&&b.b5==b5&&b.b6==b6)
            return 1;
        return 0;
    }
} sai,tem,tel;
int dice[10][10][10][10][10][10];
int a[10];
int bfs()
{
    queue<saizi> q;
    sai.size=0;
    q.push(sai);
    while(!q.empty())
    {
        sai=q.front();
        q.pop();
        if(sai==tel)
            return sai.size;
        if(dice[sai.b1][sai.b2][sai.b3][sai.b4][sai.b5][sai.b6]==1)
            continue;
        dice[sai.b1][sai.b2][sai.b3][sai.b4][sai.b5][sai.b6]=1;
        tem=saizi(sai.b6,sai.b5,sai.b3,sai.b4,sai.b1,sai.b2,sai.size+1);
        q.push(tem);
        tem=saizi(sai.b5,sai.b6,sai.b3,sai.b4,sai.b2,sai.b1,sai.size+1);
        q.push(tem);
        tem=saizi(sai.b3,sai.b4,sai.b2,sai.b1,sai.b5,sai.b6,sai.size+1);
        q.push(tem);
        tem=saizi(sai.b4,sai.b3,sai.b1,sai.b2,sai.b5,sai.b6,sai.size+1);
        q.push(tem);
    }
    return -1;
}

int main()
{
    int ans;
    while(scanf("%d",&sai.b1)!=EOF)
    {
        memset(dice,0,sizeof(dice));

        scanf("%d",&sai.b2);
        scanf("%d",&sai.b3);
        scanf("%d",&sai.b4);
        scanf("%d",&sai.b5);
        scanf("%d",&sai.b6);

        scanf("%d",&tel.b1);
        scanf("%d",&tel.b2);
        scanf("%d",&tel.b3);
        scanf("%d",&tel.b4);
        scanf("%d",&tel.b5);
        scanf("%d",&tel.b6);

        ans=bfs();
        printf("%d\n",ans);
    }
    return 0;
}

代码如下：

#include <cstdio>
#include <cstring>
struct Dice
{
    int top, bottom;
    int left, right;
    int front_f, back_b;
    int step;
} a[100017];
int goal[6];

int bfs()
{
    int tp = 1, tl = 0;
    Dice t, f;
    a[0].step = 0;
    while(1)
    {
        t = a[tl++];
        if(t.top==goal[0]&&t.bottom==goal[1]&&t.left==goal[2]&&t.right==goal[3]&&t.front_f==goal[4]&&t.back_b==goal[5])
        {
            return t.step;
            break;
        }
        else if(t.step > 4)
        {
            return t.step = -1;
            break;
        }
        else
        {
            f.top = t.front_f;//向后
            f.bottom = t.back_b;
            f.left = t.left;
            f.right = t.right;
            f.front_f = t.bottom;
            f.back_b = t.top;
            f.step = t.step+1;
            a[tp++] = f;

            f.top = t.back_b;//向前
            f.bottom = t.front_f;
            f.left = t.left;
            f.right = t.right;
            f.front_f = t.top;
            f.back_b = t.bottom;
            f.step = t.step+1;
            a[tp++] = f;

            f.top = t.right;//向左
            f.bottom = t.left;
            f.left = t.top;
            f.right = t.bottom;
            f.front_f = t.front_f;
            f.back_b = t.back_b;
            f.step = t.step+1;
            a[tp++] = f;

            f.top = t.left;//向右
            f.bottom = t.right;
            f.left = t.bottom;
            f.right = t.top;
            f.front_f = t.front_f;
            f.back_b = t.back_b;
            f.step = t.step+1;
            a[tp++] = f;
        }
    }
}

int main()
{
    while(~scanf("%d",&a[0].top))
    {
        scanf("%d",&a[0].bottom);
        scanf("%d",&a[0].left);
        scanf("%d",&a[0].right);
        scanf("%d",&a[0].front_f);
        scanf("%d",&a[0].back_b);

        for(int i = 0; i < 6; i++)
        {
            scanf("%d",&goal[i]);
        }

        int ans = bfs();
        printf("%d\n",ans);
    }
    return 0;
}

HDU 5012 Dice (bfs + 记忆化搜索)