hdu 3335 Divisibility(Dancing Links重复覆盖)

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本文探讨了一道结合数论与图论的问题，通过使用深度优先搜索算法解决了一个关于数的整除性的最大集合选择问题。该问题要求从一组正整数中挑选尽可能多的数，但这些数之间不能相互整除。

Divisibility

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 749 Accepted Submission(s): 258

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

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.

Author

DaiZhenyang@BUPT

Source

HDOJ Monthly Contest – 2010.03.06

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题目： http://acm.hdu.edu.cn/showproblem.php?pid=3335

分析：这题话说是二分图匹配。。。表示看不出来，于是用DLX来搞了，一个n*n的01矩阵，i行j列为1表示，第i个数能被第j个数整除或整除它，于是变成求取最多次使得每一列都至少包含一个1。。。然后居然1Y= =

代码：

#include<cstdio>
using namespace std;
const int mm=1000010;
const int mn=1111;
int D[mm],U[mm],L[mm],R[mm],C[mm];
int H[mn],S[mn];
__int64 a[mn];
bool v[mn];
int n,m,size,ans;
void prepare(int r,int c)
{
    for(int i=0;i<=c;++i)
    {
        S[i]=0;
        U[i]=D[i]=i;
        L[i+1]=i;
        R[i]=i+1;
    }
    R[size=c]=0;
    while(r)H[r--]=-1;
}
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;
}
int f()
{
    int c,ret=0;
    for(c=R[0];c;c=R[c])++ret;
    return ret;
}
void Dance(int k)
{
    if(k+f()<=ans)return;
    if(!R[0])
    {
        ans=k;
        return;
    }
    int i,j,c,tmp=mm;
    for(i=R[0];i;i=R[i])
        if(S[i]<tmp)tmp=S[c=i];
    for(i=D[c];i!=c;i=D[i])
    {
        remove(i);
        for(j=R[i];j!=i;j=R[j])remove(j);
        Dance(k+1);
        for(j=L[i];j!=i;j=L[j])resume(j);
        resume(i);
    }
}
void Link(int r,int c)
{
    ++S[C[++size]=c];
    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;
    }
}
int main()
{
    int i,j,T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(i=1;i<=n;++i)scanf("%I64d",&a[i]);
        prepare(n,n);
        for(i=1;i<=n;++i)
            for(j=1;j<=n;++j)
                if((a[i]%a[j])==0||(a[j]%a[i])==0)Link(i,j);
        ans=0;
        Dance(0);
        printf("%d\n",ans);
    }
    return 0;
}