HDU 3335 Divisibility(DLX可重复覆盖）

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

131 2 3 

 

Sample Output

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

 

题意：题目是要求找相互之间没有约数的最大个数。

DLX做法：满足条件的我们bool 矩阵取值为1，取最大可重复覆盖即可解决。

#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<string>#include<iostream>#include<queue>#include<cmath>#include<map>#include<stack>#include<bitset>using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )typedef long long LL;typedef pair<int,int>pil;const int maxn = 1000+5;const int maxnnode=maxn*maxn;const int mod = 1000000007;struct DLX{    int n,m,size;    int U[maxnnode],D[maxnnode],L[maxnnode],R[maxnnode],Row[maxnnode],Col[maxnnode];    int H[maxn],S[maxn];//H[i]位置，S[i]个数    int ansd;    void init(int a,int b)    {        n=a;  m=b;        REPF(i,0,m)        {            S[i]=0;            U[i]=D[i]=i;            L[i]=i-1;            R[i]=i+1;        }        R[m]=0; L[0]=m;        size=m;        REPF(i,1,n)           H[i]=-1;    }    void link(int r,int c)    {        ++S[Col[++size]=c];        Row[size]=r;        D[size]=D[c];        U[D[c]]=size;        U[size]=c;        D[c]=size;        if(H[r]<0)  H[r]=L[size]=R[size]=size;        else        {            R[size]=R[H[r]];            L[R[H[r]]]=size;            L[size]=H[r];            R[H[r]]=size;        }    }    void remove(int c)    {        for(int i=D[c];i!=c;i=D[i])            L[R[i]]=L[i],R[L[i]]=R[i];    }    void resume(int c)    {        for(int i=U[c];i!=c;i=U[i])            L[R[i]]=R[L[i]]=i;    }    bool v[maxn];    int f()    {        int ret = 0;        for(int c = R[0];c != 0;c = R[c])v[c] = true;        for(int c = R[0];c != 0;c = R[c])            if(v[c])            {                ret++;                v[c] = false;                for(int i = D[c];i != c;i = D[i])                    for(int j = R[i];j != i;j = R[j])                        v[Col[j]] = false;            }        return ret;    }    void Dance(int d)    {//        if(d + f() >=ansd) return ;        if(R[0] == 0)        {            if(d>ansd)  ansd=d;            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])        {            remove(i);            for(int j = R[i];j != i;j = R[j])remove(j);            Dance(d+1);            for(int j = L[i];j != i;j = L[j])resume(j);            resume(i);        }    }};DLX L;LL num[maxn];int t,n;int main(){    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        L.init(n,n);        REPF(i,1,n)  scanf("%I64d",&num[i]);        REPF(i,1,n)        {            REPF(j,1,n)              if(num[i]%num[j]==0||num[j]%num[i]==0)  L.link(i,j);        }        L.ansd=0;        L.Dance(0);        printf("%d\n",L.ansd);    }    return 0;}