1 /*  2     题意：就是给定两个筛子，每个筛子上6个面，每个面的数字属于[1,6]， 且互不相同！  3     问a筛子最少经过按照题目规定的要求转动，达到和b筛子上下左右前后的数字相同！  4       5     思路：很直白的bfs，将每一种状态对应一个数字，保证这种状态不会重新加入队列中！   6 */  7 #include<iostream>  8 #include<cstdio>  9 #include<cstring> 10 #include<algorithm> 11 #include<queue> 12 using namespace std; 13  14 int a[7], ss; 15 int vis[654330]; 16  17 struct node{ 18     int k[7]; 19     node(){} 20     node(int a1, int a2, int a3, int a4, int a5, int a6){ 21         k[1]=a1; 22         k[2]=a2; 23         k[3]=a3; 24         k[4]=a4; 25         k[5]=a5; 26         k[6]=a6; 27     } 28     int step; 29 }; 30  31 queue<node>q;  32  33  34 int sum(node x){ 35     int s=0; 36     for(int i=1; i<=6; ++i) 37        s= s*10 + x.k[i]; 38     return s; 39 } 40  41 bool bfs(){ 42     while(!q.empty()) q.pop(); 43     node cur(a[1], a[2], a[3], a[4], a[5], a[6]); 44     cur.step=0; 45     q.push(cur); 46     vis[sum(cur)]=1; 47     while(!q.empty()){ 48         cur = q.front(); 49         if(sum(cur)==ss){ 50             printf("%d\n", cur.step); 51             return true; 52         } 53         q.pop(); 54         node *nt = new node(cur.k[5], cur.k[6], cur.k[3], cur.k[4], cur.k[2], cur.k[1]); 55         int v = sum(*nt); 56         if(!vis[v]){ 57             vis[v]=1; 58             nt->step = cur.step + 1;  59             q.push(*nt); 60         } 61          62         nt = new node(cur.k[6], cur.k[5], cur.k[3], cur.k[4], cur.k[1], cur.k[2]); 63         v = sum(*nt); 64         if(!vis[v]){ 65             vis[v]=1; 66             nt->step = cur.step + 1;  67             q.push(*nt); 68         } 69          70         nt = new node(cur.k[3], cur.k[4], cur.k[2], cur.k[1], cur.k[5], cur.k[6]); 71         v = sum(*nt); 72         if(!vis[v]){ 73             vis[v]=1; 74             nt->step = cur.step + 1;  75             q.push(*nt); 76         } 77          78         nt = new node(cur.k[4], cur.k[3], cur.k[1], cur.k[2], cur.k[5], cur.k[6]); 79         v = sum(*nt); 80         if(!vis[v]){ 81             vis[v]=1; 82             nt->step = cur.step + 1;  83             q.push(*nt); 84         } 85     } 86     return false; 87 } 88  89 int main(){ 90     while(scanf("%d%d%d%d%d%d", &a[1], &a[2], &a[3], &a[4], &a[5], &a[6])!=EOF){ 91         ss=0; 92         for(int i=1; i<=6; ++i){ 93             int x; 94             scanf("%d", &x); 95             ss = ss * 10 + x; 96         } 97         memset(vis, 0, sizeof(vis)); 98         if(!bfs()) 99             printf("-1\n"); 100     }101     return 0;102 }