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 }