1. 程式人生 > >BZOJ 3101(N皇后-N皇后O(n)構造一組解的方法)

BZOJ 3101(N皇后-N皇后O(n)構造一組解的方法)

3101: N皇后

Time Limit: 10 Sec  Memory Limit: 128 MBSec  Special Judge
Submit: 70  Solved: 32
[Submit][Status]

Description

n*n的棋盤,在上面擺下n個皇后,使其兩兩間不能相互攻擊…

Input

一個數n

Output

i行表示在第i行第幾列放置皇后

Sample Input

4

Sample Output

2
4
1
3



HINT

100%的資料3<n<1000000。輸出任意一種合法解即可

Source



以下是找到的N皇后一組解得構造法:

一、當n mod 6 != 2 或 n mod 6 != 3時,有一個解為:
2,4,6,8,...,n,1,3,5,7,...,n-1 (n為偶數)
2,4,6,8,...,n-1,1,3,5,7,...,n (n為奇數)
(上面序列第i個數為ai,表示在第i行ai列放一個皇后;... 省略的序列中,相鄰兩數以2遞增。下同)
二、當n mod 6 == 2 或 n mod 6 == 3時,
(當n為偶數,k=n/2;當n為奇數,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為偶數,n為偶數)
k,k+2,k+4,...,n-1,2,4,...,k-2,k+3,k+5,...,n-2,1,3,5,...,k+1,n (k為偶數,n為奇數)
k,k+2,k+4,...,n-1,1,3,5,...,k-2,k+3,...,n,2,4,...,k+1 (k為奇數,n為偶數)
k,k+2,k+4,...,n-2,1,3,5,...,k-2,k+3,...,n-1,2,4,...,k+1,n (k為奇數,n為奇數)

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Forstep(i,n,step) for(int i=1;i<=n;i+=step)
#define Forkstep(i,k,n,step) for(int i=k;i<=n;i+=step)
  
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (1000000+10)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
int a[MAXN]; 
int main()
{
//	freopen("bzoj3101.in","r",stdin);
//	freopen(".out","w",stdout);
	int n,j=0;
	cin>>n;
	if (n%6!=2&&n%6!=3)
	{
		For(i,n) if (i%2==0) a[++j]=i;
		For(i,n) if (i%2==1) a[++j]=i;
		
	} 
	else if (n%6==2) 
	{	
		int k=n>>1;
		for(int i=k;i<=n;i+=2) a[++j]=i;
		for(int i=k%2?1:2;i<=k-2;i+=2) a[++j]=i;
		for(int i=k+3;i<=n;i+=2) a[++j]=i;
		for(int i=k%2?2:1;i<=k+1;i+=2) a[++j]=i;
	}
	else if (n%6==3) 
	{	
		int k=n>>1;
		for(int i=k;i<n;i+=2) a[++j]=i;
		for(int i=k%2?1:2;i<=k-2;i+=2) a[++j]=i;
		for(int i=k+3;i<n;i+=2) a[++j]=i;
		for(int i=k%2?2:1;i<=k+1;i+=2) a[++j]=i;
		a[++j]=n;
	}
	
	For(i,n) printf("%d\n",a[i]);
	
	return 0;
}