我写了一个实现来解决C ++中的骑士之旅，即使对于像8这样的小值，它似乎也花费了太长时间。我使用的是大多数书籍中所示的基本回溯方法，并且in_board和unvisited函数检查是否移动制作在木板间内，到现在为止都没有访问。

移动向量包含如下对

    {-1,2},{-1,-2},{1,{2,-1},1},{-2,

bool
knights_tour(vector < vector <int> > &board,int x,int y,int d)
{
        int new_x,new_y;
        if (d == board.size()*board.size())
        {
                print_board(board,"Here is a solution");
                return true;
        }


        for (auto move : moves)
        {
                new_x = x + move.first;
                new_y = y + move.second;

                if (in_board(board,new_x,new_y) && unvisited(board,new_y))
                {
                        board[new_x][new_y] = d;

                        if (knights_tour(board,new_y,d+1))
                                return true;
                        else
                                board[new_x][new_y] = -1;
                }
        }
        return false;
}




int main()
{
        int n;
        cin >> n;
        vector <vector <int>> board(n,vector <int> (n,-1));
        board[0][0] = 0;
        cout << board.size()*board.size();
        knights_tour(board,1);
}