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

 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 }