八皇后问题

在8X8格的国际象棋上摆放八个皇后，使其不能互相攻击，即任意两个皇后都不能处于同一行、同一列或同一斜线上，问有多少种摆法。试解出92种结果。

// eight_queen.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"

#include<deque>
#include<iostream>


using namespace std;


#define N 8


typedef unsigned char BYTE;


BYTE chess_board[N][N] = { 0 };
deque<deque<BYTE>>record, solution;

deque<BYTE>cc;
deque<BYTE>bb;


bool update(int height)
{
	cc.clear();
	chess_board[height][bb[height]] = 1;
	for (int i = 0; i < N; i++)
	{
		chess_board[height][i] = 1;
		chess_board[i][bb[height]] = 1;
		if (height - bb[height] + i >= 0 && height - bb[height] + i <= N - 1)
			chess_board[height - bb[height] + i][i] = 1;
		if (height + bb[height] - i >= 0 && height + bb[height] - i <= N - 1)//右上左下
			chess_board[height + bb[height] - i][i] = 1;
	}

	for (int i = 0; i < N; i++)
		if (chess_board[height + 1][i] == 0)
			cc.push_back(i);
	if (!cc.empty())
	{
		return true;
	}
	return false;
}




void eight_queen()
{

	for (int i = 0; i < N; i++)
	{
		bb.push_back(i);
	}
	record.push_back(bb);
	bb.clear();
	int k = 0;

	while (k < N)
	{
		if (!record.empty())
		{
			if (record[k].empty())
			{
				while (record[k].empty())
				{
					record.pop_back();
					if (record.empty())
						return;
					bb.pop_back();
					k--;
				}
				for (int i = 0; i < N; i++)
					for (int j = 0; j < N; j++)
						chess_board[i][j] = 0;
				for (int i = 0; i < k; i++)
					update(i);
			}
			bb.push_back(record[k][0]);
			record[k].pop_front();
		}

		if (!update(k))
		{
			while (!update(k))
			{
				
				if (record[k].empty())
				{
					while (record[k].empty())
					{
						record.pop_back();
						if (record.empty())
							return;
						k--;
						for (int i = 0; i < N; i++)
							for (int j = 0; j < N; j++)
								chess_board[i][j] = 0;
						for (int i = 0; i < k; i++)
							update(i);
						bb.pop_back();
					}
					bb.pop_back();
					break;
				}

				else
				{

					bb[k] = record[k][0];
					record[k].pop_front();
					for (int i = 0; i < N; i++)
						for (int j = 0; j < N; j++)
							chess_board[i][j] = 0;
					for (int i = 0; i < k; i++)
						update(i);

					if (update(k))
					{
						if (k == N - 2)
						{
							deque<BYTE> dd;
							dd = bb;
							dd.push_back(0);
							for (int i = 0; i < cc.size(); i++)
							{
								dd[N - 1] = cc[i];
								solution.push_back(dd);
							}
							
							bb.pop_back();
							for (int i = 0; i < N; i++)
								for (int j = 0; j < N; j++)
									chess_board[i][j] = 0;
							for (int i = 0; i < k; i++)
								update(i);
						}
						else
						{
							record.push_back(cc);
							k++;
						}
						break;
					}
				}

			}
		}
		else
		{
			if (k == N - 2)
			{
				deque<BYTE> dd;
				dd = bb;
				dd.push_back(0);
				for (int i = 0; i < cc.size(); i++)
				{
					dd[N - 1] = cc[i];
					solution.push_back(dd);
				}
			
				bb.pop_back();
				for (int i = 0; i < N; i++)
					for (int j = 0; j < N; j++)
						chess_board[i][j] = 0;
				for (int i = 0; i < k; i++)
					update(i);
			}
			else
			{
				record.push_back(cc);
				k++;
			}
		}
	}


}

int _tmain(int argc, _TCHAR* argv[])
{
	eight_queen();

	system("pause");
	return 0;
}

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