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 8typedef 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;}