用的BFS，将每一个数字除以4，得到四种不同的状态：0 1 2 3，为了实现判重，用2位表示一个状态，整个钟表需要2*9 = 18位，不大；
题解中第二种做法貌似是推理得到了一般解法。
1 /* 2 PROG : clocks 3 LANG : C++ 4 */ 5 6 # include <cstdio> 7 # include <cstring> 8 # include <queue> 9 10 using namespace std; 11 12 int ss; 13 char vis[0x3FFFF + 0x1]; 14 int pre[0x3FFFF + 0x1]; 15 int solu[200]; 16 17 /*********************************/ 18 const int od[][6] = 19 { 20 {0,1,3,4, -1,0}, 21 {0,1,2,-1,0,0}, 22 {1,2,4,5,-1,0}, 23 {0,3,6,-1,0,0}, 24 {1,3,4,5,7,-1}, 25 {2,5,8,-1,0,0}, 26 {3,4,6,7,-1,0}, 27 {6,7,8,-1,0,0}, 28 {4,5,7,8,-1,0} 29 }; 30 /*********************************/ 31 32 int flip(int s, int id) 33 { 34 int i, t, tt; 35 for (i = 0; od[id][i] != -1; ++i) 36 { 37 //printf("\t%d\t", od[id][i]); 38 t = 2 * od[id][i]; 39 tt = ((((s>>t)&0x3)+1)&0x3); 40 s &= ~(0x3<<t); 41 s |= tt<<t; 42 } 43 return s; 44 } 45 46 void bfs(int ss) 47 { 48 int x, y, i; 49 queue <int> Q; 50 51 if (ss == 0x3FFFF) return ; 52 53 memset(vis, 0, sizeof(vis)); 54 while (!Q.empty()) Q.pop(); 55 56 vis[ss] = 1; 57 Q.push(ss); 58 while (!Q.empty()) 59 { 60 x = Q.front(); Q.pop(); 61 for (i = 0; i < 9; ++i) 62 { 63 y = flip(x, i); 64 //printf("%05X\t%d\n", y, i+1); 65 if (!vis[y]) 66 { 67 vis[y] = i+1; 68 pre[y] = x; 69 if (y == 0x3FFFF) 70 return ; 71 Q.push(y); 72 } 73 } 74 } 75 } 76 77 void print(int s) 78 { 79 if (s == ss) return ; 80 print(pre[s]); 81 printf("%d", vis[s]); 82 if (s != 0x3FFFF) putchar(' '); 83 else putchar('\n'); 84 } 85 86 void init(void) 87 { 88 int i, x; 89 for (ss = i = 0; i < 9; ++i) 90 { 91 scanf("%d", &x); 92 ss |= ((x>>2)<<(i*2)); 93 } 94 //printf("%0X\n", ss); 95 } 96 97 int main() 98 { 99 freopen("clocks.in", "r", stdin); 100 freopen("clocks.out", "w", stdout); 101 102 init(); 103 bfs(ss); 104 print(0x3FFFF); 105 106 fclose(stdin); 107 fclose(stdout); 108 109 return 0; 110 }