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