C#实现八皇后问题，有点看不懂，能帮忙分析下吗

Question

http://www.cnblogs.com/ziyiFly/archive/2008/09/10/1288097.html 部分解析

Answer 1

bool GetNewQueen(int size, int step)
        {
            bool Metch = false;
            string resultStr = string.Empty;            for (int i = 0; i < size; i++)
            {
                Metch = false;
                Queen[step] = i;                if (step > 0)
                {
                    for (int j = 0; j < step; j++)
                    {
                        int d = Queen[j];
                        if (i == d || i == d + j - step || i == d - j + step)
                        {
                            Metch = true;
                            break;
                        }
                    }                    if (!Metch)
                    {
                        if (step < size-1)
                        {
                            Metch = GetNewQueen(size, step + 1);
                        }
                        else
                        {
                            for (int k = 0; k < size; k++)
                            {
                                resultStr += Queen[k] + ",";
                            }
                            richTextBox1.Text += resultStr + "\n";
                        }
                    }
                }
                else
                {
                    Metch = GetNewQueen(size, step + 1);
                }
            }调用:
int[ ] Queen;
void Main()
{
int size=8;
Queen=new int[size];
GetNewQueen(size,0);
}

Answer 2

已测试.
size=2或3,无结果
size=4,结果2
size=5,结果10
size=8,结果92

Answer 3

晕,少粘贴2行代码后面是
return Metch;
}

Answer 4

八皇后算法实现
  //皇后数量
        static void QueenArithmetic(int size)
        {
            int[] Queen = new int[size];//每行皇后的位置
            int y, x, i, j, d, t = 0;
            y = 0;
            Queen[0] = -1;
            while (true)
            {
                for (x = Queen[y] + 1; x < size; x++)
                {
                    for (i = 0; i < y; i++)
                    {
                        j = Queen[i];
                        d = y - i;
                        //检查新皇后是否能与以前的皇后相互攻击
                        if ((j == x) || (j == x - d) || (j == x + d))
                            break;
                    }
                    if (i >= y)
                        break;//不攻击
                }
                if (x == size) //没有合适的位置
                {
                    if (0 == y)
                    {
                        //回溯到了第一行
                        Console.WriteLine("Over");
                        break; //结束
                    }//CodeGo.net/
                    //回溯
                    Queen[y] = -1;
                    y--;
                }
                else
                {
                    Queen[y] = x;//确定皇后的位置
                    y++;//下一个皇后
                    if (y < size)
                        Queen[y] = -1;
                    else
                    {
                        //所有的皇后都排完了，输出
                        Console.WriteLine("\n" + ++t + ':');
                        for (i = 0; i < size; i++)
                        {
                            for (j = 0; j < size; j++)
                                Console.Write(Queen[i] == j ? 'Q' : '*');
                            Console.WriteLine();
                        }
                        y = size - 1;//回溯
                    }
                }
            }
            Console.ReadLine();
        }
        #endregion        static void Main(string[] args)
        {
            int size = 8;//皇后数
            QueenArithmetic(size);
        }