POJ 3239 Solution to the n Queens Puzzle 犀利利的构造公式

 

Solution to the n Queens Puzzle

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 2897		Accepted: 1067		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 arbitraryn ×n boards. Given n, you are to find a solution to then queens puzzle.

Input

The input contains multiple test cases. Each test case consists of a single integern 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 theith 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

 

题意：在N*N的上 摆下N个皇后 使得所有的皇后都不会被其他皇后攻击；

           皇后 可以 攻击到 同行 列 ，斜线；

思路：不知道怎么构造，只会DFS 回溯

             但是N的范围到了300，太大了  DFS效率太低。

             网上看到牛人想出来的 构造公式 好强大了 ORZ、

/*
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)
*/

#include <iostream>
using namespace std;
int main()
{
	int i,k,n;
	while(scanf("%d", &n), n)
	{
		if(n%6!=2 && n%6!=3)
		{
			printf("2");
			for(i=4; i<=n; i+=2)
				printf(" %d", i);
			for(i=1; i<=n; i+=2)
				printf(" %d", i);
			printf("\n");
			continue;
		}
		k = n/2;
		printf("%d", k);
		if(!(k & 1) && !(n & 1))
		{
			for(i=k+2; i<=n; i+=2)
				printf(" %d", i);
			for(i=2; i<=k-2; i+=2)
				printf(" %d", i);
			for(i=k+3; i<=n-1; i+=2)
				printf(" %d", i);
			for(i=1; i<=k+1; i+=2)
				printf(" %d", i);        
		}
		else if(!(k & 1) && (n & 1))
		{
			for(i=k+2; i<=n-1; i+=2)
				printf(" %d", i);
			for(i=2; i<=k-2; i+=2)
				printf(" %d", i);
			for(i=k+3; i<=n-2; i+=2)
				printf(" %d", i);            
			for(i=1; i<=k+1; i+=2)
				printf(" %d", i);
			printf(" %d", n);        
		}
		else if((k & 1) && !(n & 1))
		{
			for(i=k+2; i<=n-1; i+=2)
				printf(" %d", i);
			for(i=1; i<=k-2; i+=2)
				printf(" %d", i);
			for(i=k+3; i<=n; i+=2)
				printf(" %d", i);
			for(i=2; i<=k+1; i+=2)
				printf(" %d", i);
		}
		else if((k&1) && (n&1))
		{
			for(i=k+2; i<=n-2; i+=2)
				printf(" %d", i);
			for(i=1; i<=k-2; i+=2)
				printf(" %d", i);
			for(i=k+3; i<=n-1; i+=2)
				printf(" %d", i);
			for(i=2; i<=k+1; i+=2)
				printf(" %d", i);
			printf(" %d", n);
		}
		printf("\n");
	}
	return 0;
}