• poj3239


    题意:给出正方形棋盘边长n(最大300),要求输出一种摆放n皇后不冲突的方案。

    分析:

    数据范围较大,只能用构造的方法,不能用搜索。

    下面用一个数列表示一种方案,第i个数表示棋盘第i行上的皇后所在的列号

    n皇后问题构造法:

    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。

    View Code
    #include <iostream>
    #include
    <cstdio>
    #include
    <cstdlib>
    #include
    <cstring>
    usingnamespace std;

    int n;

    int main()
    {
    //freopen("t.txt", "r", stdin);
    while (scanf("%d", &n), n)
    {
    if (n %6!=2&& n %6!=3)
    {
    printf(
    "2");
    for (int i =4; i <= n; i +=2)
    printf(
    " %d", i);
    for (int i =1; i <= n; i +=2)
    printf(
    " %d", i);
    putchar(
    '\n');
    continue;
    }
    int k = n /2;
    printf(
    "%d", k);
    if (!(k &1) &&!(n &1))
    {
    for (int i = k +2; i <= n; i +=2)
    printf(
    " %d", i);
    for (int i =2; i <= k -2; i +=2)
    printf(
    " %d", i);
    for (int i = k +3; i <= n -1; i +=2)
    printf(
    " %d", i);
    for (int i =1; i <= k +1; i +=2)
    printf(
    " %d", i);
    }
    elseif (!(k &1) && (n &1))
    {
    for (int i = k +2; i <= n -1; i +=2)
    printf(
    " %d", i);
    for (int i =2; i <= k -2; i +=2)
    printf(
    " %d", i);
    for (int i = k +3; i <= n -2; i +=2)
    printf(
    " %d", i);
    for (int i =1; i <= k +1; i +=2)
    printf(
    " %d", i);
    printf(
    " %d", n);
    }
    elseif ((k &1) &&!(n &1))
    {
    for (int i = k +2; i <= n -1; i +=2)
    printf(
    " %d", i);
    for (int i =1; i <= k -2; i +=2)
    printf(
    " %d", i);
    for (int i = k +3; i <= n; i +=2)
    printf(
    " %d", i);
    for (int i =2; i <= k +1; i +=2)
    printf(
    " %d", i);
    }
    elseif ((k &1) && (n &1))
    {
    for (int i = k +2; i <= n -2; i +=2)
    printf(
    " %d", i);
    for (int i =1; i <= k -2; i +=2)
    printf(
    " %d", i);
    for (int i = k +3; i <= n -1; i +=2)
    printf(
    " %d", i);
    for (int i =2; i <= k +1; i +=2)
    printf(
    " %d", i);
    printf(
    " %d", n);
    }
    putchar(
    '\n');
    }
    return0;
    }

     

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  • 原文地址:https://www.cnblogs.com/rainydays/p/2104336.html
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