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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
80Sample Output
5 3 1 6 8 2 4 7Source
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
1 /*代码一：构造法*/ 2 3 //Memory Time 4 //188K 16MS 5 6 #include<iostream> 7 #include<cmath> 8 using namespace std; 9 10 int main(int i)11 {12 int n; //皇后数13 while(cin>>n)14 {15 if(!n)16 break;17 18 if(n%6!=2 && n%6!=3)19 {20 if(n%2==0) //n为偶数21 {22 for(i=2;i<=n;i+=2)23 cout<<i<<‘ ‘;24 for(i=1;i<=n-1;i+=2)25 cout<<i<<‘ ‘;26 cout<<endl;27 }28 else //n为奇数29 {30 for(i=2;i<=n-1;i+=2)31 cout<<i<<‘ ‘;32 for(i=1;i<=n;i+=2)33 cout<<i<<‘ ‘;34 cout<<endl;35 }36 }37 else if(n%6==2 || n%6==3)38 {39 if(n%2==0) //n为偶数40 {41 int k=n/2;42 if(k%2==0) //k为偶数43 {44 for(i=k;i<=n;i+=2)45 cout<<i<<‘ ‘;46 for(i=2;i<=k-2;i+=2)47 cout<<i<<‘ ‘;48 for(i=k+3;i<=n-1;i+=2)49 cout<<i<<‘ ‘;50 for(i=1;i<=k+1;i+=2)51 cout<<i<<‘ ‘;52 cout<<endl;53 }54 else //k为奇数55 {56 for(i=k;i<=n-1;i+=2)57 cout<<i<<‘ ‘;58 for(i=1;i<=k-2;i+=2)59 cout<<i<<‘ ‘;60 for(i=k+3;i<=n;i+=2)61 cout<<i<<‘ ‘;62 for(i=2;i<=k+1;i+=2)63 cout<<i<<‘ ‘;64 cout<<endl;65 }66 }67 else //n为奇数68 {69 int k=(n-1)/2;70 if(k%2==0) //k为偶数71 {72 for(i=k;i<=n-1;i+=2)73 cout<<i<<‘ ‘;74 for(i=2;i<=k-2;i+=2)75 cout<<i<<‘ ‘;76 for(i=k+3;i<=n-2;i+=2)77 cout<<i<<‘ ‘;78 for(i=1;i<=k+1;i+=2)79 cout<<i<<‘ ‘;80 cout<<n<<endl;81 }82 else //k为奇数83 {84 for(i=k;i<=n-2;i+=2)85 cout<<i<<‘ ‘;86 for(i=1;i<=k-2;i+=2)87 cout<<i<<‘ ‘;88 for(i=k+3;i<=n-1;i+=2)89 cout<<i<<‘ ‘;90 for(i=2;i<=k+1;i+=2)91 cout<<i<<‘ ‘;92 cout<<n<<endl;93 }94 }95 }96 }97 return 0;98 }
Poj 3239 Solution to the n Queens Puzzle