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.
我原先觉得01矩阵行和列也是给出的每个数，但是精确覆盖不可能啊，重复覆盖的话，像样例那样选择1和2也是可以的，但是和题意不相符，1和2是不可以被同时选的。。。。
但是题里面只要求个数，这样找出来并不影响答案的正确性。。。。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<queue>
#include<cmath>
using namespace std;
typedef long long LL;
const int inf = 0x3f3f3f3f3f;
const double eps = 1e-8;
const double PI = acos(-1);
#define pb push_back
#define mp make_pair
#define fi first
#define se second
//最大行数
const int MN = 1005;
//最大列数
const int MM = 1005;
//最大点数
const int MNN = 1e5 + 5 + MM;
struct DLX
{
//一共n行m列，s个节点
int n,m,s;
//交叉十字链表组成部分
//第i个节点的上U下D左L右R，所在位置row行col列
int U[MNN],D[MNN],L[MNN],R[MNN],row[MNN],col[MNN];
//H数组记录行选择指针，S数组记录覆盖个数
int H[MN],S[MM];
//res记录行个数，ans数组记录可行解
int res,ans[MN];
//初始化空表
void init(int x,int y)
{
n = x,m = y;
//其中0节点作为head节点,其他作为列首节点
for(int i = 0;i <= m;++i){
U[i] = D[i] = i;
L[i] = i - 1;
R[i] = i + 1;
}
R[m] = 0;L[0] = m;
s = m;
memset(S,0,sizeof(S));
memset(H,-1,sizeof(H));
}
void Insert(int r,int c)
{
//节点数加一，设置s节点所处位置，以及S列覆盖个数加一
s++;row[s] = r;col[s] = c;S[c]++;
//将s节点插入对应列中
D[s] = D[c];U[D[c]] = s;
U[s] = c;D[c] = s;
if(H[r] < 0){//如果该行没有元素，H[r]标记该行起始节点
H[r] = L[s] = R[s] = s;
}else{
//将该节点插入该行第一个节点后面
R[s] = R[H[r]];
L[R[H[r]]] = s;
L[s] = H[r];
R[H[r]] = s;
}
}
//精确覆盖
void Remove(int c)
{
//删除c列
L[R[c]] = L[c];R[L[c]] = R[c];
//删除该列上的元素对应的行
for(int i = D[c];i != c;i = D[i]){//枚举该列元素
for(int j = R[i];j != i;j = R[j]){//枚举列的某个元素所在行遍历
U[D[j]] = U[j];
D[U[j]] = D[j];
//将该列上的S数组减一
--S[col[j]];
}
}
}
void resume(int c)
{
//恢复c列
for(int i = U[c];i != c;i = U[i]){//枚举该列元素
for(int j = L[i];j != i;j = L[j]){
U[D[j]] = j;D[U[j]] = j;
++S[col[j]];
}
}
L[R[c]] = c;R[L[c]] = c;
}
bool dance(int deep)
{
if(res < deep) return false;
//当矩阵为空时,说明找到一个可行解，算法终止
if(R[0] == 0){
res = min(res,deep);
return true;
}
//找到节点数最少的列，枚举这列上的所有行
int c = R[0];
for(int i = R[0];i != 0;i = R[i]){
if(S[i] < S[c]){
c = i;
}
}
//删除节点数最少的列
Remove(c);
for(int i = D[c];i != c;i = D[i]){
//将行r放入当前解
ans[deep] = row[i];
//行上节点对应的列上进行删除
for(int j = R[i];j != i;j = R[j])
Remove(col[j]);
//进入下一层
dance(deep + 1);
//对行上的节点对应的列进行恢复
for(int j = L[i];j != i;j = L[j])
resume(col[j]);
}
//恢复节点数最少列
resume(c);
return false;
}
//重复覆盖
//将列与矩阵完全分开
void Remove1(int c)
{
for(int i = D[c];i != c;i = D[i]){
L[R[i]] = L[i];
R[L[i]] = R[i];
}
}
void resume1(int c)
{
for(int i = D[c];i != c;i = D[i]){
L[R[i]] = R[L[i]] = i;
}
}
int vis[MNN];
//估价函数，模拟删除列，H()，函数返回的是至少还需要多少行才能完成重复覆盖
int A()
{
int dis = 0;
for(int i = R[0];i != 0;i = R[i]) vis[i] = 0;
for(int i = R[0];i != 0;i = R[i]){
if(!vis[i]){
dis++;vis[i] = 1;
for(int j = D[i];j != i;j = D[j]){
for(int k = R[j];k != j;k = R[k]){
vis[col[k]] = 1;
}
}
}
}
return dis;
}
void dfs(int deep)
{
if(!R[0]){
//cout << res << endl;
res = max(res,deep);
return ;
}
//if(deep + A() >= res) return ;
int c = R[0];
for(int i = R[0];i != 0;i = R[i]){
if(S[i] < S[c]){
c = i;
}
}
for(int i = D[c];i != c;i = D[i]){
//每次将第i列其他节点删除，只保留第i节点，为了找该行的节点
Remove1(i);
//将列上的节点完全与矩阵脱离，只删列首节点是不行的
for(int j = R[i];j != i;j = R[j]){
Remove1(j);
}
dfs(deep + 1);
for(int j = L[i];j != i;j = L[j]){
resume1(j);
}
resume1(i);
}
}
}dlx;
const int N = 1005;
LL num[N];
int n;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
dlx.init(n,n);
for(int i = 1;i <= n;++i)
scanf("%lld",&num[i]);
for(int i = 1;i <= n;++i){
for(int j = 1;j <= n;++j){
if(num[j] % num[i] == 0 || num[i] % num[j] == 0){
//cout << i << " " << j << endl;
dlx.Insert(i,j);
}
}
}
dlx.res = 0;
dlx.dfs(0);
printf("%d\n",dlx.res);
}
return 0;
}