您的位置:首页 > 其它

八皇后问题

2015-07-29 02:01 337 查看
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

= { 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;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: