HDU 3335 Divisibility

Problem Description

As we know,the fzu AekdyCoin is famous of math,especially in the field of number theory.So,many people call him "the descendant of Chen Jingrun",which brings him a good reputation.

AekdyCoin also plays an important role in the ACM_DIY group,many people always ask him questions about number theory.One day,all members urged him to conduct a lesson in the group.The rookie daizhenyang is extremely weak at math,so he is delighted.

However,when AekdyCoin tells us "As we know, some numbers have interesting property. For example, any even number has the property that could be divided by 2.",daizhenyang got confused,for he don't have the concept of divisibility.He asks other people for help,first,he randomizely writes some positive integer numbers,then you have to pick some numbers from the group,the only constraint is that if you choose number a,you can't choose a number divides a or a number divided by a.(to illustrate the concept of divisibility),and you have to choose as many numbers as you can.

Poor daizhenyang does well in neither math nor programming.The responsibility comes to you!

Input

An integer t,indicating the number of testcases,

For every case, first a number n indicating daizhenyang has writen n numbers(n<=1000),then n numbers,all in the range of (1...2^63-1).

Output

The most number you can choose.

Sample Input

1

3

1 2 3

Sample Output

2

Hint:

If we choose 2 and 3,one is not divisible by the other,which is the most number you can choose.

dlx重复覆盖的巧用，重复覆盖的原理恰好符合本题的筛选方式，即选择一个数后，该数的倍数或约数可以保证在之后的搜索中不会被选择

于是修改一下启发函数，求解最大的重复覆盖即可。

#include<cstdio>
#include<vector>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll maxn = 5005;
int T, n, m, x, y, t, X, Y,tot;
ll a[maxn];

inline void read(int &ret)
{
	char c;
	do {
		c = getchar();
	} while (c < '0' || c > '9');
	ret = c - '0';
	while ((c = getchar()) >= '0' && c <= '9')
		ret = ret * 10 + (c - '0');
}

struct DLX
{
	#define maxn 500005
	#define F(i,A,s) for (int i=A[s];i!=s;i=A[i])
	int L[maxn], R[maxn], U[maxn], D[maxn];
	int row[maxn], col[maxn], ans[maxn], cnt[maxn];
	int n, m, num, sz;

	void add(int now, int l, int r, int u, int d, int x, int y)
	{
		L[now] = l;	R[now] = r;	U[now] = u;
		D[now] = d;   row[now] = x;  col[now] = y;
	}
	void reset(int n, int m)
	{
		num = 0x7FFFFFFF;
		this->n = n;	this->m = m;
		for (int i = 0; i <= m; i++)
		{
			add(i, i - 1, i + 1, i, i, 0, i);
			cnt[i] = 0;
		}
		L[0] = m; 	R[m] = 0; 	sz = m + 1;
	}
	void insert(int x, int y)
	{
		int ft = sz - 1;
		if (row[ft] != x)
		{
			add(sz, sz, sz, U[y], y, x, y);
			U[D[sz]] = sz; D[U[sz]] = sz;
		}
		else
		{
			add(sz, ft, R[ft], U[y], y, x, y);
			R[L[sz]] = sz; L[R[sz]] = sz;
			U[D[sz]] = sz; D[U[sz]] = sz;
		}
		++cnt[y];	++sz;
	}

	//精确覆盖
	void remove(int now)
	{
		R[L[now]] = R[now];
		L[R[now]] = L[now];
		F(i, D, now) F(j, R, i)
		{
			D[U[j]] = D[j];
			U[D[j]] = U[j];
			--cnt[col[j]];
		}
	}
	void resume(int now)
	{
		F(i, U, now)	F(j, L, i)
		{
			D[U[j]] = j;
			U[D[j]] = j;
			++cnt[col[j]];
		}
		R[L[now]] = now;
		L[R[now]] = now;
	}
	bool dfs(int x)
	{
		//if (x + A() >= num) return;
		if (x==n) { num = min(num, x); return true; }
		int now = R[0];
		if (now>n) return false;
		for (int i=R[0];i<=n;i=R[i]) if (cnt[now]>cnt[i]) now = i;
		remove(now);
		F(i, D, now)
		{
			ans[x] = row[i];
			F(j, R, i) remove(col[j]);
			if (dfs(x + 1)) return true;
			F(j, L, i) resume(col[j]);
		}
		resume(now);
		return false;
	}
	//精确覆盖

	//重复覆盖
	void Remove(int now)
	{
		F(i, D, now)
		{
			L[R[i]] = L[i];
			R[L[i]] = R[i];
		}
	}
	void Resume(int now)
	{
		F(i, U, now) L[R[i]] = R[L[i]] = i;
	}
	int vis[maxn];
	int flag[maxn];
	int A()
	{
		int dis = 0;
		F(i, R, 0) dis++;
		return dis;
	}
	void Dfs(int x)
	{
		if (!R[0]) num = max(num, x);
		else if (A()+x>num)
		{
			int now = R[0];
			F(i, R, 0) if (cnt[now]>cnt[i]) now = i;
			F(i, D, now)
			{
				Remove(i); F(j, R, i) Remove(j);
				Dfs(x + 1);
				F(j, L, i) Resume(j); Resume(i);
			}
		}
	}
	//重复覆盖

}dlx;

int main()
{
	read(T);
	while (T--)
	{
		scanf("%d", &n);
		dlx.reset(n,n);
		for (int i=1;i<=n;i++) scanf("%lld",&a[i]);
		for (int i=1;i<=n;i++)
			for (int j=1;j<=n;j++)
				if (a[i]%a[j]==0||a[j]%a[i]==0)
					dlx.insert(i,j);
		dlx.num=0;
		dlx.Dfs(0);
		printf("%d\n",dlx.num);
	}
	return 0;
}