LeetCode N-Queens

class Solution {
private:
    int queen_num;
    vector<vector<string> > res;
public:
    vector<vector<string> > solveNQueens(int n) {
        res.clear();
        queen_num = n;
        
        vector<bool> h(n, false);
        vector<bool> v(n, false);
        vector<bool> lx(n * 2 + 1, false);
        vector<bool> rx(n * 2 + 1, false);

        vector<int> solution;
        
        dfs(solution, h, v, lx, rx);
        
        return res;
    }

    void dfs(vector<int> &mem, vector<bool> &h, vector<bool> &v, vector<bool> &lx, vector<bool> &rx) {
        if (mem.size() == queen_num) {
            add_solution(mem);
            return;
        }
        int row = mem.size();
        int col = 0;
        while (col < queen_num) {
            if (!h[row] && !v[col] && !lx[row + col] && !rx[row + queen_num - col - 1]) {
                h[row] = v[col] = lx[row + col] = rx[row + queen_num - col - 1] = true;
                mem.push_back(col);
                dfs(mem, h, v, lx, rx);
                mem.pop_back();
                h[row] = v[col] = lx[row + col] = rx[row + queen_num - col - 1] = false;
            }
            col++;
        }
    }

    void add_solution(vector<int> &solution) {
        res.push_back(vector<string>());

        string line(queen_num, '.');
        for (int i=0; i<queen_num; i++) {
            line[solution[i]] = 'Q';
            res.back().push_back(line);
            line[solution[i]] = '.';
        }
    }
};

第二轮：

The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.

Given an integer n, return all distinct solutions to the n-queens puzzle.

Each solution contains a distinct board configuration of the n-queens' placement, where 'Q' and '.' both indicate a queen and an empty space respectively.

For example,
There exist two distinct solutions to the 4-queens puzzle:

[
 [".Q..",  // Solution 1
  "...Q",
  "Q...",
  "..Q."],

 ["..Q.",  // Solution 2
  "Q...",
  "...Q",
  ".Q.."]
]

class Solution {
private:
    vector<vector<string> > solutions;
public:
    vector<vector<string> > solveNQueens(int n) {
        vector<int> H(n, 0);
        vector<int> V(n, 0);
        vector<int> S1(n*2, 0);
        vector<int> S2(n*2, 0);
        
        vector<string> res;
        
        solutions.clear();
        
        dfs(n, 0, H, V, S1, S2, res);
        
        return solutions;
    }
    
    void dfs(int n, int i, vector<int>& H, vector<int>& V, vector<int>& S1, vector<int>& S2, vector<string>& res) {
        if (i == n) {
            solutions.push_back(res);
            return;
        }
        for (int j=0; j<n; j++) {
            if (!H[i] && !V[j] && !S1[n-i-1 + j] && !S2[i+j]) {
                H[i] = V[j] = S1[n-i-1 + j] = S2[i+j] = 1;
                string s(n, '.');
                s[j] = 'Q';
                res.push_back(s);
                dfs(n, i + 1, H, V, S1, S2, res);
                res.pop_back();
                H[i] = V[j] = S1[n-i-1 + j] = S2[i+j] = 0;
            }
        }
    }
};

对于行可以直接用0, 1, 2...n这些数字表示，数字代表某行上Queen所在的列，我们只需要把这些预先准备好的行做一个全排序，得到的结果在行列上肯定不会又冲突，此时只要检查对角线是否冲突即可：

class Solution {
private:
    vector<vector<string> > res;
public:
    /**
     * Get all distinct N-Queen solutions
     * @param n: The number of queens
     * @return: All distinct solutions
     * For example, A string '...Q' shows a queen on forth position
     */
    vector<vector<string> > solveNQueens(int n) {
        // write your code here
        res.clear();
        
        vector<int> m(n);
        vector<int> vd1(2 * n - 1, false);
        vector<int> vd2(2 * n - 1, false);
        
        for (int i = 0; i < n; i++) {
            m[i] = i;
        }
        
        dfs(n, 0, m, vd1, vd2);
        
        return res;
    }
    
    void dfs(int n, int pos, vector<int>& m, vector<int>& vd1, vector<int>& vd2) {
        if (pos == n) {
            vector<string> r(n, string(n, '.'));
            for (int i = 0; i < n; i++) {
                r[i][m[i]] = 'Q';
            }
            res.push_back(r);
            return;
        }
        for (int i = pos; i < n; i++) {
            swap(m[pos], m[i]);
            int col = m[pos];
            int row = pos;
            int& v1 = vd1[row + col];
            int& v2 = vd2[n - row - 1 + col];
            if (!v1 && !v2) {
                v1 = v2 = true;
                dfs(n, pos + 1, m, vd1, vd2);
                v1 = v2 = false;
            }
            swap(m[pos], m[i]);
        }
    }

};