Poj 3239 Solution to the n Queens Puzzle

1.Link：

http://poj.org/problem?id=3239

2.Content：

Solution to the n Queens Puzzle

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 3459		Accepted: 1273		Special Judge

Description

The eight queens puzzle is the problem of putting eight chess queens on an 8 × 8 chessboard such that none of them is able to capture any other. The puzzle has been generalized to arbitrary n × n boards. Given n, you are to find a solution to the n queens puzzle.

Input

The input contains multiple test cases. Each test case consists of a single integer n between 8 and 300 (inclusive). A zero indicates the end of input.

Output

For each test case, output your solution on one line. The solution is a permutation of {1, 2, …, n}. The number in the ith place means the ith-column queen in placed in the row with that number.

Sample Input

8
0

Sample Output

5 3 1 6 8 2 4 7

Source

POJ Monthly--2007.06.03, Yao, Jinyu

3.Method:

一开始用8皇后的方法，发现算不出来。

只能通过搜索，可以利用构造法，自己也想不出来构造，所以直接套用了别人的构造公式

感觉没啥意义，直接就用别人的代码提交了，也算是完成一道题目了

构造方法：

http://www.cnblogs.com/rainydays/archive/2011/07/12/2104336.html

一、当n mod 6 != 2 且 n mod 6 != 3时，有一个解为：
2,4,6,8,...,n,1,3,5,7,...,n-1        (n为偶数)
2,4,6,8,...,n-1,1,3,5,7,...,n        (n为奇数)
(上面序列第i个数为ai,表示在第i行ai列放一个皇后;...省略的序列中,相邻两数以2递增。下同)
二、当n mod 6 == 2 或 n mod 6 == 3时，
(当n为偶数,k=n/2；当n为奇数,k=(n-1)/2)
k,k+2,k+4,...,n,2,4,...,k-2,k+3,k+5,...,n-1,1,3,5,...,k+1        (k为偶数,n为偶数)
k,k+2,k+4,...,n-1,2,4,...,k-2,k+3,k+5,...,n-2,1,3,5,...,k+1,n    (k为偶数,n为奇数)
k,k+2,k+4,...,n-1,1,3,5,...,k-2,k+3,...,n,2,4,...,k+1            (k为奇数,n为偶数)
k,k+2,k+4,...,n-2,1,3,5,...,k-2,k+3,...,n-1,2,4,...,k+1,n        (k为奇数,n为奇数)

第二种情况可以认为是，当n为奇数时用最后一个棋子占据最后一行的最后一个位置，然后用n-1个棋子去填充n-1的棋盘，这样就转化为了相同类型且n为偶数的问题。

若k为奇数，则数列的前半部分均为奇数，否则前半部分均为偶数。

4.Code：

http://blog.csdn.net/lyy289065406/article/details/6642789?reload

/*代码一：构造法*/

//Memory Time 
//188K   16MS 

#include<iostream>
#include<cmath>
using namespace std;

int main(int i)
{
    int n;  //皇后数
    while(cin>>n)
    {
        if(!n)
            break;

        if(n%6!=2 && n%6!=3)
        {
            if(n%2==0)  //n为偶数
            {
                for(i=2;i<=n;i+=2)
                    cout<<i<<' ';
                for(i=1;i<=n-1;i+=2)
                    cout<<i<<' ';
                cout<<endl;
            }
            else   //n为奇数
            {
                for(i=2;i<=n-1;i+=2)
                    cout<<i<<' ';
                for(i=1;i<=n;i+=2)
                    cout<<i<<' ';
                cout<<endl;
            }
        }
        else if(n%6==2 || n%6==3)
        {
            if(n%2==0)  //n为偶数
            {
                int k=n/2;
                if(k%2==0)  //k为偶数
                {
                    for(i=k;i<=n;i+=2)
                        cout<<i<<' ';
                    for(i=2;i<=k-2;i+=2)
                        cout<<i<<' ';
                    for(i=k+3;i<=n-1;i+=2)
                        cout<<i<<' ';
                    for(i=1;i<=k+1;i+=2)
                        cout<<i<<' ';
                    cout<<endl;
                }
                else  //k为奇数
                {
                    for(i=k;i<=n-1;i+=2)
                        cout<<i<<' ';
                    for(i=1;i<=k-2;i+=2)
                        cout<<i<<' ';
                    for(i=k+3;i<=n;i+=2)
                        cout<<i<<' ';
                    for(i=2;i<=k+1;i+=2)
                        cout<<i<<' ';
                    cout<<endl;
                }
            }
            else   //n为奇数
            {
                int k=(n-1)/2;
                if(k%2==0)  //k为偶数
                {
                    for(i=k;i<=n-1;i+=2)
                        cout<<i<<' ';
                    for(i=2;i<=k-2;i+=2)
                        cout<<i<<' ';
                    for(i=k+3;i<=n-2;i+=2)
                        cout<<i<<' ';
                    for(i=1;i<=k+1;i+=2)
                        cout<<i<<' ';
                    cout<<n<<endl;
                }
                else  //k为奇数
                {
                    for(i=k;i<=n-2;i+=2)
                        cout<<i<<' ';
                    for(i=1;i<=k-2;i+=2)
                        cout<<i<<' ';
                    for(i=k+3;i<=n-1;i+=2)
                        cout<<i<<' ';
                    for(i=2;i<=k+1;i+=2)
                        cout<<i<<' ';
                    cout<<n<<endl;
                }
            }
        }
    }
    return 0;
}