C++ 用回溯法解决N皇后问题

N皇后问题是计算机科学中最为经典的问题之一。1848年，国际西洋棋起手马克思·贝瑟尔提出了8皇后问题。
将N个皇后摆放在N*N的棋盘中，互相不可攻击，有多少种摆放方式，每种摆放方式具体是怎样的？

#include<vector>
#include<string>
class Solution
{
public:
 Solution() {};
 ~Solution() {};
 std::vector<std::vector<std::string>> solveNQueens(int n)
 {
  std::vector<std::vector<std::string>> result;
  std::vector<std::vector<int>> mark;
  std::vector<std::string> location;
  for (int i = 0; i < n; i++)
  {
   mark.push_back((std::vector<int>()));
   for (int j = 0; j < n; j++)
   {
    mark[i].push_back(0);
   }
   location.push_back("");
   location[i].append(n,'.');
  }
  generate(0,n,location,mark,result);
  return result;
 }
private:
 void put_down_the_queen(int x, int y, std::vector<std::vector<int>>& mark)
 {
  static const int dx[] = { -1,1,0,0,-1,-1,1,1 };
  static const int dy[] = { 0,0,-1,1,-1,1,-1,1 };
  mark[x][y] = 1;
  for (unsigned int i = 1; i < mark.size(); i++)
  {
   for (int j = 0; j < 8; j++)
   {
    int new_x = x + i * dx[j];
    int new_y = y + i * dy[j];
    if (new_x >= 0 && new_x < mark.size() && new_y >= 0 && new_y < mark.size())
    {
     mark[new_x][new_y] = 1;
    }
   }
  }
 }
 void generate(int k, int n, std::vector<std::string>& location, std::vector<std::vector<int>> &mark,  std::vector<std::vector<std::string>> &result)
 {
  if (k==n)
  {
   result.push_back(location);
   return;
  }
  for (int i = 0; i < n; i++)
  {
   if (mark[k][i]==0)
   {
    std::vector<std::vector<int>> tmp_mark=mark;
    location[k][i] = 'Q';
    put_down_the_queen(k, i, mark);
    generate(k + 1, n, location, mark, result);
    mark = tmp_mark;
    location[k][i] = '.';
   }
  }
 }
};
int main()
{
 std::vector<std::vector<std::string>> result;
 Solution solve;
 result = solve.solveNQueens(4);
 for (unsigned int i = 0; i < result.size(); i++)
 {
  printf("i=%d\n",i);
  for (int j = 0; j < result[i].size(); j++)
  {
   printf("%s\n",result[i][j].c_str());
  }
  printf("\n");
 }
 return 0;
}

运行结果为：

i=0
.Q..
...Q
Q...
..Q.

i=1
..Q.
Q...
...Q
.Q..