*蓝桥OJ 1508N皇后问题 DFS

 蓝桥OJ 1508N皇后问题

//首先我们要理清限制条件
//皇后所在行列不能有重复皇后,且其对角线也不能
 
#include
using namespace std;
int n, ans;
const int N = 15;
int vis[N][N];
 
void dfs(int dep)
{
  //如果dep==n+1,说明此放置方法合理
  if(dep == n+1)
  {
    ans++;
    return;
  }
 
  for (int i = 1; i <= n; i++)
  {
    if (vis[dep][i]) continue;
 
  //根据dep可以判断每行只会放置1个皇后
  //所以只需要把每列和对角线排除
    for (int _i = 1; _i <= n; _i++) vis[_i][i]++;
    for (int _i = dep,_j = i; _i >= 1 & _j <= n; --_i,++_j) vis[_i][_j]++;
    for (int _i = dep,_j = i; _i <= n & _j >= 1; ++_i,--_j) vis[_i][_j]++;
    for (int _i = dep,_j = i; _i >= 1 & _j >= 1; --_i,--_j) vis[_i][_j]++;
    for (int _i = dep,_j = i; _i <= n & _j <= n; ++_i,++_j) vis[_i][_j]++;
 
  dfs(dep+1);
 
  //恢复
    for (int _i = 1; _i <= n; _i++) vis[_i][i]--;
    for (int _i = dep,_j = i; _i >= 1 & _j <= n; --_i,++_j) vis[_i][_j]--;
    for (int _i = dep,_j = i; _i <= n & _j >= 1; ++_i,--_j) vis[_i][_j]--;
    for (int _i = dep,_j = i; _i >= 1 & _j >= 1; --_i,--_j) vis[_i][_j]--;
    for (int _i = dep,_j = i; _i <= n & _j <= n; ++_i,++_j) vis[_i][_j]--;
 
  }
}
 
int main()
{
  ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
  cin >> n;
  dfs(1);
  cout << ans << '\n';
  return 0;
}