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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

    8
    0

    Sample Output

    5 3 1 6 8 2 4 7

    Source

    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 }
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  • 原文地址:https://www.cnblogs.com/mobileliker/p/4056949.html
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