皇后问题之C#版（非递归）

/*
 *author:junyi sun @ccnu
* e-mail:[email protected]
*/

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