C + + recursive implementation n Queens Problem code (eight Queens problem) _c language

Let's take a look at the basics of the 8 Queens question:

Put eight queens on 8x8 's chess, so that they can't attack each other, that is, any two queens can not be in the same row, the same column or the same slash, ask how many kinds of pendulum method.

Extending to n Queens is the same problem.
Look, it seems to be using a two-dimensional array. Actually, it's not necessary. One-dimensional arrays can be judged, such as arr[i], to represent an element in row I arr[i] column--a wide range of tips. And here we do not have to consider to store the entire matrix, if Arr[i] exists, then we print, print to the Queen position when the output of 1, non-queen bit output 0 can be.

This way of thinking of the realization of a lot of online, including the previous mention of the classmate, so also do not have to struggle to improve there is no improvement and so on, the right to be a practice.

Directly on the code well, feel that the recursive method is nothing to say, space to think of the ability to better understand a little bit easily. Tomorrow is free to write and write not recursive implementation bar.

Copy Code code as follows:

/*
* NQueen.cpp
*
* Created on:2013 Year December 23
* Author:nerohwang
*/
The formal parameter rowcurrent represents the number of rows currently in the
#include <iostream>
#include <fstream>
#include <iomanip>
#include <stdlib.h>
using namespace Std;
BOOL Check (int rowcurrent,int *&nqueen); Judgment function
void Print (Ofstream &os,int n,int *&nqueen); Print function
void Solve (int rowcurrent,int *&nqueen,int n,int &count, Ofstream &os); n Queen problem handling function, index general initial value is 0

//Judgment function, either there is a conflict at all, or there is a conflict on the slash, return false
bool Check (int rowcurrent,int *&nqueen)
{
    int i = 0;
    while (i < rowcurrent)
    {
         if (nqueen[i] = = Nqueen[rowcurrent] | | (ABS (nqueen[i]-nqueen[rowcurrent]) = = ABS (i-rowcurrent))
        {
             return false;
       }
        i++;
   }
    return true;
}

//Output text document for all possible results
void Print (ofstream &os,int n,int *&nqueen)
{
    os<< " One call \ n ";
    for (int i = 0;i < n;i++) {
        for (int j = 0; J < N J + +)
        {
             os<< (nqueen[i]==j?1:0);
            OS<<SETW (2);
       }
        os<< "\ n";
   }
    os<< "\ n";
}

//core functions. Recursively solves the N-Queens problem and prints the bottom of the
void Solve (int rowcurrent,int *&nqueen,int n,int &count, ofstream &os)
{
    if (rowcurrent = n)  //Current line number bottom, that is, complete a matrix, output it
    {
         Print (Os,n,nqueen);
        count++;
   }
    for (int i = 0;  i < n; i++)
    {
     & nbsp;  Nqueen[rowcurrent] = i;                     //row Row I column try
        if check ( Rowcurrent,nqueen))
        {
             Solve (Rowcurrent+1,nqueen,n,count,os); //Move Down line
        }
   }

int main ()
{
int n; Scale of problems
int count = 0; Count of Solutions
cout<< "Please enter the size of the problem n" <<endl;
cin>>n;
if (n<4)
{
cerr<< "problem scale must be greater than 4" <<endl;
return 0;
}
int *nqueen = new Int[n];
Ofstream OS;
Os.open ("Result.txt");
Solve (0,nqueen,n,count,os);
cout<< "The solution of the problem is" <<count<< "Way" <<endl;
Os.close ();
return 0;
}