很容易判断是BFS，可是，呵呵呵呵呵呵。。。。。。。。。
HASH判重吧，判连通可以用并查集。
以下代码是转别人的，我码了一下午，发觉越码越丑，呵呵了。
http://www.cnblogs.com/Lyush/p/3416507.html
#include <cstdlib> #include <cstring> #include <cstdio> #include <iostream> #include <algorithm> #include <queue> using namespace std; // 9! = 362880 const int LIM = 362880; int fac[9]; // 递增进制数的权值 int tot[4]; // 四种颜色的个数 char vis[LIM]; // hash九宫格 char op[6]; // 滚动的操作的限制 char hav[50]; // 4*9个格子的标记 struct Block { char color[6]; int id; }bk[3][3]; struct Node { Block *ptr[3][3]; int key; int ti; Node() {} Node(Block (*x)[3]) { ti = 0; for (int i = 0; i < 3; ++i) { for (int j = 0; j < 3; ++j) { ptr[i][j] = &x[i][j]; } } getkey(); } void show() { for (int i = 0; i < 3; ++i) { for (int j = 0; j < 3; ++j) { printf("%5d", ptr[i][j]->id); } puts(""); } puts(""); } void getkey() { int cnt; key = 0; for (int i = 0; i < 8; ++i) { cnt = 0; for (int j = i+1; j < 9; ++j) { if (ptr[j/3][j%3]->id < ptr[i/3][i%3]->id) ++cnt; } key += cnt * fac[8-i]; } } int dfs(int x, int y, int z) { if (hav[(x*3+y)*4+z]) return 0; hav[(x*3+y)*4+z] = 1; char ch = ptr[x][y]->color[z]; int ret = 1; if (z == 0) { if (ptr[x][y]->color[2] == ch) ret += dfs(x, y, 2); if (ptr[x][y]->color[3] == ch) ret += dfs(x, y, 3); if (x > 0 && ptr[x-1][y]->color[1] == ch) ret += dfs(x-1, y, 1); } else if (z == 1) { if (ptr[x][y]->color[2] == ch) ret += dfs(x, y, 2); if (ptr[x][y]->color[3] == ch) ret += dfs(x, y, 3); if (x < 2 && ptr[x+1][y]->color[0] == ch) ret += dfs(x+1, y, 0); } else if (z == 2) { if (ptr[x][y]->color[0] == ch) ret += dfs(x, y, 0); if (ptr[x][y]->color[1] == ch) ret += dfs(x, y, 1); if (y > 0 && ptr[x][y-1]->color[3] == ch) ret += dfs(x, y-1, 3); } else { if (ptr[x][y]->color[0] == ch) ret += dfs(x, y, 0); if (ptr[x][y]->color[1] == ch) ret += dfs(x, y, 1); if (y < 2 && ptr[x][y+1]->color[2] == ch) ret += dfs(x, y+1, 2); } return ret; } bool judge() { memset(hav, 0, sizeof (hav)); for (int i = 0; i < 3; ++i) { for (int j = 0; j < 3; ++j) { for (int k = 0; k < 4; ++k) { if (!hav[(i*3+j)*4+k]) { int cnt = dfs(i, j, k); switch(ptr[i][j]->color[k]) { case 'R': if (cnt != tot[0]) return false; break; case 'G': if (cnt != tot[1]) return false; break; case 'B': if (cnt != tot[2]) return false; break; default : if (cnt != tot[3]) return false; } } } } } return true; } }; void pre() { fac[1] = 1; for (int i = 2; i < 9; ++i) { fac[i] = fac[i-1] * i; } } void swapr(Block *x[], Block *y[], int dir) { // 横向的滚动 if (dir == 0) { // 向左 y[0] = x[1], y[1] = x[2], y[2] = x[0]; } else { // 向右 y[0] = x[2], y[1] = x[0], y[2] = x[1]; } } void swapc(Block *x[], Block *y[], int dir) { // 纵向的滚动 if (dir == 0) { // 向上 y[0] = x[3], y[3] = x[6], y[6] = x[0]; } else { // 向下 y[0] = x[6], y[3] = x[0], y[6] = x[3]; } } int bfs() { queue<Node>q; Node tmp = Node(bk); Node pos; memset(vis, 0, sizeof (vis)); q.push(tmp); // 初始化是没有逆序对的 vis[tmp.key] = 1; while (!q.empty()) { pos = q.front(); q.pop(); if (pos.judge()) { return pos.ti; } for (int i = 0; i < 6; ++i) { if (op[i]) continue; tmp = pos; if (i < 3) { // 横向的滚动 for (int j = 0; j < 2; ++j) { swapr(&pos.ptr[i][0], &tmp.ptr[i][0], j); tmp.getkey(); tmp.ti = pos.ti + 1; if (!vis[tmp.key]) { vis[tmp.key] = 1; q.push(tmp); } } } else { // 纵向的 for (int j = 0; j < 2; ++j) { swapc(&pos.ptr[0][i%3], &tmp.ptr[0][i%3], j); tmp.getkey(); tmp.ti = pos.ti + 1; if (!vis[tmp.key]) { vis[tmp.key] = 1; q.push(tmp); } } } } } } int main() { int T, ca = 0; pre(); scanf("%d", &T); while (T--) { memset(op, 0, sizeof (op)); memset(tot, 0, sizeof (tot)); for (int i = 0; i < 3; ++i) { for (int j = 0; j < 3; ++j) { scanf("%s", bk[i][j].color); for (int k = 0; k < 4; ++k) { switch(bk[i][j].color[k]) { case 'R': ++tot[0]; break; case 'G': ++tot[1]; break; case 'B': ++tot[2]; break; default : ++tot[3]; } } bk[i][j].id = i*3+j; // 0-8 if (bk[i][j].color[4] == '1') { op[i] = op[3+j] = 1; // 两种操作无法进行 } } } printf("Case #%d: %d ", ++ca, bfs()); } return 0; }