皇后问题

/*
 *Author:Junyi Sun @CCNU
* E-mail:[e-mail]
*/

using System;
namespace sunjoy
{
    public class Queen
    {
        public static int Main()
        {
            int board_size = 0,x=0,y=0;//棋盘大小,当前行,当前列
            uint solution_count = 0;   //皇后摆放方案的个数
            int[] rows, cols, slot1, slot2, x2y;//行占用情况，列占用情况，“/”状斜线占用情况，“\”状斜线占用情况，皇后坐标
            DateTime t_start, t_end;
           
            Console.WriteLine("请输入棋盘的大小");
            try
            {
                board_size = Convert.ToInt32(Console.ReadLine());
                if (board_size <= 0)
                {
                    Console.WriteLine("非法数据");
                    return -1;
                }
            }
            catch (Exception e) { Console.WriteLine("发生异常" + e.Message); };

            rows = new int[board_size];
            cols = new int[board_size];
            slot1 = new int[board_size * 2 - 1];
            slot2 = new int[board_size * 2 - 1];
            x2y = new int[board_size];

            for (int i = 0; i < board_size; i++)
                x2y[i] = -1;   //坐标初始化都为-1
            t_start = DateTime.Now;
            while (true)
            {
                for (y = x2y[x]+1; y < board_size; y++)
                    if (rows[x] == 0 && cols[y] == 0 && slot1[x + y] == 0 && slot2[x - y + board_size - 1] == 0)
                        break;

                if (y < board_size)
                {
                    //第X行的棋子落下
                    rows[x] = 1; cols[y] = 1; slot1[x + y] = 1; slot2[x - y + board_size - 1] = 1; x2y[x] = y;
                }  
                else
                {
                    //回溯，拿起棋子
                    if (x > 0)
                    {
                        x2y[x] = -1;
                        x--;
                        rows[x] = 0; cols[x2y[x]] = 0; slot1[x + x2y[x]] = 0; slot2[x - x2y[x] + board_size - 1] = 0;
                        continue;
                    }
                    else break;
                }
                if (x == board_size - 1)   //如果已经得到了一组解，即当最后一行棋子落定之时
                {
                    for (int i = 0; i < board_size; i++)
                    {
                        for (int j = 0; j < board_size; j++)
                        {
                            if (x2y[i] == j) Console.Write("Ｑ");
                            else Console.Write("■");
                        }
                        Console.Write("\n");
                       
                    }
                    Console.Write("\n");
                    solution_count++;   //总方案数加一
                    rows[x] = 0; cols[x2y[x]] = 0; slot1[x + x2y[x]] = 0; slot2[x - x2y[x] + board_size - 1] = 0;//放弃这一列

                }
                else
                {
                    x++;   //继续处理下一行
                }
            }
            t_end = DateTime.Now;
            Console.WriteLine("总共{0}组解",solution_count);
            Console.WriteLine("计算及打印共用时间{0}秒",t_end-t_start);
            return 0;
        }
    }
}

___________________________________________________________

测试结果:
请输入棋盘的大小
4
■Ｑ■■
■■■Ｑ
Ｑ■■■
■■Ｑ■

■■Ｑ■
Ｑ■■■
■■■Ｑ
■Ｑ■■

总共2组解
计算及打印共用时间00:00:00秒

_______________________________________________
请输入棋盘的大小
6
■Ｑ■■■■
■■■Ｑ■■
■■■■■Ｑ
Ｑ■■■■■
■■Ｑ■■■
■■■■Ｑ■

■■Ｑ■■■
■■■■■Ｑ
■Ｑ■■■■
■■■■Ｑ■
Ｑ■■■■■
■■■Ｑ■■

■■■Ｑ■■
Ｑ■■■■■
■■■■Ｑ■
■Ｑ■■■■
■■■■■Ｑ
■■Ｑ■■■

■■■■Ｑ■
■■Ｑ■■■
Ｑ■■■■■
■■■■■Ｑ
■■■Ｑ■■
■Ｑ■■■■

总共4组解
计算及打印共用时间00:00:00.0156250秒