读懂题意以后还很容易做的，

和类似加一个维度，筛子的形态，可以用上方的点数u和前面的点数f来表示，相对的面点数之和为7，可以预先存储u和f的对应右边的点数，点数转化就很容易了。

具体做法看代码

#include
     
      using namespace std;const int maxn = 11;const int maxd = 7;int g[maxn][maxn];bool vis[maxn][maxn][maxd][maxd];struct State{    int x,y,u,f,pre;}Q[6000];int tor[maxd][maxd];inline void trans(int u,int f,int i,int &nu,int &nf){    if(i == 0){ nu = f; nf = 7-u; return; }    if(i == 1){ nu = 7 - tor[u][f]; nf = f; return; }    if(i == 2){ nu = 7 - f; nf = u ; return; }    nu = tor[u][f]; nf = f;}vector
      
        ans;void print_ans(int rst){    for(int i = rst ; ~i; i = Q[i].pre){        ans.push_back(i);    }    vector
       
        ::reverse_iterator ri = ans.rbegin(),ed = ans.rend();    int cnt = 1;    for( ed--; ri != ed; ri++,cnt++){        printf("(%d,%d),",Q[*ri].x,Q[*ri].y);        if(!(cnt%9)) printf("\n  ");    }    printf("(%d,%d)\n",Q[*ed].x,Q[*ed].y);}void bfs(int sx,int sy,int su,int sf){    const int dx[] = {-1,0,1,0};    const int dy[] = {
   0,1,0,-1};    int head,rear;    head = rear = 0;    Q[rear].u = su; Q[rear].f = sf; Q[rear].x = sx; Q[rear].y = sy; Q[rear].pre = -1;    rear++;    while(head