参考了http://blog.csdn.net/y1196645376/article/details/69718192，这个大哥的思路很巧妙。
思路：
dfs。
实现：
1 #include <iostream> 2 #include <cstdio> 3 using namespace std; 4 5 const int N = 6; 6 const int dx[4] = { 0, 1, -1, 0 }; 7 const int dy[4] = { 1, 0, 0, -1 }; 8 9 bool vis[6][6]; 10 11 int dfs(int x, int y) 12 { 13 if (x == 0 || x == N || y == 0 || y == N) 14 { 15 return 1; 16 } 17 int cnt = 0; 18 for (int i = 0; i < 4; i++) 19 { 20 int nx = x + dx[i]; 21 int ny = y + dy[i]; 22 if (nx >= 0 && nx <= N && ny >= 0 && ny <= N && !vis[nx][ny]) 23 { 24 vis[nx][ny] = true; 25 vis[N - nx][N - ny] = true; 26 cnt += dfs(nx, ny); 27 vis[nx][ny] = false; 28 vis[N - nx][N - ny] = false; 29 } 30 } 31 return cnt; 32 } 33 34 int main() 35 { 36 vis[3][3] = true; 37 cout << dfs(3, 3) / 4 << endl; 38 return 0; 39 }