Leetcode:N-Queens

最新推荐文章于 2025-09-09 22:06:54 发布

转载最新推荐文章于 2025-09-09 22:06:54 发布 · 65 阅读

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原文链接：http://www.cnblogs.com/Kai-Xing/p/3883561.html

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#数据结构与算法

本文介绍了一种使用深度优先搜索(DFS)与回溯法解决N皇后问题的有效方法。通过记录已使用的列、主对角线及副对角线状态，确保放置的皇后不会互相攻击。文中提供了一个C++实现示例，展示了如何找到所有不冲突的N皇后解决方案。

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.."]
]

分析：DFS+backtracking。用三个数组分别用过的列、主对角线、辅对角线。代码如下：

class Solution {
public:
    vector<bool> used_col;
    vector<bool> used_diag;
    vector<bool> used_antidiag;
    
    vector<vector<string> > solveNQueens(int n) {
        vector<vector<string> > result;
        vector<string> path;
        if(n == 0) return result;
        
        this->used_col = vector<bool>(n, false);
        this->used_diag = vector<bool>(2*n-1, false);
        this->used_antidiag = vector<bool>(2*n-1, false);
        
        dfs(result, path, n, 0);
        
        return result;
    }
    
    void dfs(vector<vector<string> > &result, vector<string> &path, int n, int row){
        if(row == n){
            result.push_back(path);
            return;
        }
        for(int i = 0; i < n; i++){
            int diag = row-i+n;
            int antidiag = row + i;
            if(!used_col[i] && !used_diag[diag] && !used_antidiag[antidiag]){
                used_col[i] = true;
                used_diag[diag] = true;
                used_antidiag[antidiag] = true;
                path.push_back(string(i,'.')+"Q"+string(n-i-1, '.'));
                dfs(result, path, n, row+1);
                path.pop_back();
                used_col[i] = false;
                used_diag[diag] = false;
                used_antidiag[antidiag] = false;
            }
        }
    }
};