Divisibility HDU

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!

InputAn 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). 
OutputThe 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.

题意：给定一组数，要求数中选取互不为约束的数，问最多选取多少个；

思路：
虽然这个题目让我们求得是最大值，其实是在最小值里面的最大值，DLX可重复覆盖的模板是为了找到最小值，所以找到最小值的过程中默认了选择某一个之后和他相关连的都不用在选择了。。。
而这样正好就符合了这个题目的要求，构造的话就是n行n列就好，代表这n个数，将与此数有约数关系的数取值为1.当选中一列时，与此列中的数有约数关系的数
都将删去。另外加一点剪枝使代码更高效。

代码：

#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <iostream>
#define mod 1000000007
#define eps 1e-6
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;

const int maxn=1010*1010;
struct DLX
{
    int left[maxn],right[maxn],up[maxn],down[maxn],col[maxn];
    int head[maxn],num[1010];
    int ansed,id;
    void init(int nn)
    {
        int n=nn;
        for(int i=0;i<=n;i++)
        {
            left[i]=i-1;
            right[i]=i+1;
            up[i]=down[i]=i;
            col[i]=i;
        }
        left[0]=n;
        right[n]=0;
        ansed=0;
        id=n;
        memset(num,0,sizeof(num));
        memset(head,-1,sizeof(head));
    }
    void link(int x,int y)
    {
        id++;
        up[id]=up[y];
        down[id]=y;
        down[up[y]]=id;
        up[y]=id;
        num[y]++;
        col[id]=y;
        if(head[x]==-1)
        {
            head[x]=left[id]=right[id]=id;
        }
        else
        {
            left[id]=left[head[x]];
            right[id]=head[x];
            right[left[head[x]]]=id;
            left[head[x]]=id;
        }
    }
    void remove(int c)//移除列
    {
        for(int i=down[c];i!=c;i=down[i])
        {
            left[right[i]]=left[i];
            right[left[i]]=right[i];
        }
    }
    void reback(int c)//恢复列
    {
        for(int i=up[c];i!=c;i=up[i])
        {
            left[right[i]]=i;
            right[left[i]]=i;
        }
    }
    int A()//未剪掉的列
    {
        int s=0;
        for(int i=right[0];i!=0;i=right[i])
        {
            s++;
        }
        return s;
    }
    void danc(int step)
    {
        if(right[0]==0)
        {
            if(step>ansed)//解中的最大值
            {
                ansed=step;
            }
            return ;
        }
        if(A()+step<=ansed)//如果剩下的步骤与已走的步骤比当前答案少，则停止循环
        {
            return ;
        }
        int c=right[0];
        for(int i=c;i!=0;i=right[i])//优先处理列中数字少的
        {
            if(num[i]<num[c])
            {
                c=i;
            }
        }
        for(int i=down[c];i!=c;i=down[i])
        {
            remove(i);
            for(int j=right[i];j!=i;j=right[j])
            {
                remove(j);
            }
            danc(step+1);
            for(int j=left[i];j!=i;j=left[j])
            {
                reback(j);
            }
            reback(i);
        }
    }
}dlx;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        ll num[1010];
        int n;
        scanf("%d",&n);
        dlx.init(n);
        for(int i=0;i<n;i++)
        {
            scanf("%lld",&num[i]);
        }
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(num[i]%num[j]==0||num[j]%num[i]==0)
                {
                    dlx.link(i+1,j+1);
                }
            }
        }
        dlx.danc(0);
        printf("%d
",dlx.ansed);
    }
}