八皇后算法 非递归 非栈

 

#include <cstdlib>
#include <iostream>
#include <string>
#include <iterator>
#include <vector>
#include <algorithm>
#include <fstream>
#include <functional>
#include <sstream>
using namespace std;

//输出一个序列
template<typename Iter>
void print(Iter first,Iter last,
                        const char* nm = "",
                        const char* sep = "/t",
                        ostream& os = cout)
{
        if(nm != 0 && *nm != '/0')
                os << nm << ": " << sep;
        typedef typename iterator_traits<Iter>::value_type VT;
        copy(first,last,ostream_iterator<VT >(os,sep));
        os << endl;
}
//八皇后算法
void eightQueen()
{
        short num = 0;
        ofstream outf("eight_queen.txt"); //结果保存在eight_queen.txt中
        short qp[9] = {0}; //qp[1]~qp[8]代表棋盘的8列，qp[i]=j表示第i列第j行放一个皇后
        for ( short lie = 1; lie <= 8; ++lie)
        {
                short hang = 0;
                for ( hang = qp[lie]+1; hang <= 8; ++hang)//此处用qp[lie]代替了栈的使用
                {
                        bool isOk = true;
                        //检查当前列当前行可否放皇后
                        for (short i = 1; i < lie; ++i)
                                if ( qp[i] == hang || abs(qp[i]-hang) == abs(i-lie) )
                                {
                                        isOk = false;
                                        break;
                                }
                        if ( isOk )
                        {//当前位置可用
                                qp[lie] = hang;
                                if ( lie == 8 )
                                {
                                        //如果是最后一列，表示搜索到一种放法，将其输出
                                        ostringstream os;
                                        os << ++num;
                                        print(qp+1,qp+9,os.str().c_str(),"/t",outf);
                                }
                                else        //否则跳出当前列，进入下一列
                                        break;
                        }
                }
                if ( hang == 9 )
                {//当前列已经搜索完，回退到上一列
                        qp[lie] = 0; //将当前列清零，保证下一次搜索该列时从第1行开始
                        lie -= 2; //for语句还要进行+1操作，减2才是回退1列
                }
                if ( lie == -1)
                        break; //第一列已经搜索完毕，算法结束
        }
}

int main(int argc, char *argv[])
{
        eightQueen();
        system("pause");
}