dfs又一经典例题，用全排列的思路解决

import java.util.*;

public class Main {
    static int n;
    static int N = 10;
    static boolean[] col = new boolean[N]; // 每一列
    static boolean[] dg = new boolean[N * 2]; // 对角线、反对角线个数会多一些
    static boolean[] udg = new boolean[N * 2]; // 反对角线
    static char[][] g = new char[N][N]; // 存储棋盘

    private static void dfs(int u) { // 以全排列的思路想n皇后问题
        if (u == n) {
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) System.out.print(g[i][j]);
                System.out.println();
            }
            System.out.println();
        } else {
            for (int i = 0; i < n; i++) {
                if (!col[i] && !dg[u + i] && !udg[n - u + i - 1]) { // 只有在该列、对角线、斜对角线都没有皇后时，才放上去
                    g[u][i] = 'Q';
                    col[i] = true;
                    dg[u + i] = true;
                    udg[n - u + i - 1] = true;
                    dfs(u + 1);
                    // 恢复现场
                    col[i] = false;
                    dg[u + i] = false;
                    udg[n - u + i - 1] = false;
                    g[u][i] = '.';
                }
            }
        }
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        n = sc.nextInt();

        for (int i = 0; i < n; i++) 
            for (int j = 0; j < n; j++)
                g[i][j] = '.';

        dfs(0);
    }
}