HDU 3335 Divisibility (DLX)

Divisibility

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 3335

Appoint description:  System Crawler  (2015-04-10)

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.

 

 

 

错了好久啊，DLX理解还是不够深。移除的时候当前行不要移除，要保持联系这样才能左右移动，还加了个剪枝函数进去，若剩下的列加上已取的列小于答案值那么就不再取了。

#include <iostream>
#include <string>
#include <cstring>
#include <cstdio>
using    namespace    std;

const    int    HEAD = 0;
const    int    SIZE = 1005 * 1005;

int    N,ANS;
int    U[SIZE],D[SIZE],L[SIZE],R[SIZE],C[1005];

void    ini(void);
void    dancing(int);
void    remove(int);
void    resume(int);
int    cut(void);
void    debug(int);
int    main(void)
{
    int        t;
    long    long    s[1005];

    //freopen("txt.txt","r",stdin);
    scanf("%d",&t);
    while(t --)
    {
        scanf("%d",&N);
        for(int i = 1;i <= N;i ++)
            scanf("%lld",&s[i]);

        ini();
        int    count = N + 1;
        for(int i = 1;i <= N;i ++)
        {
            int    first = count;
            for(int j = 1;j <= N;j ++)
                if(s[i] % s[j] == 0 || s[j] % s[i] == 0)
                {
                    U[count] = U[j];
                    D[count] = j;
                    L[count] = count - 1;
                    R[count] = count + 1;

                    D[U[count]] = count;
                    U[j] = count;
                    C[count] = j;

                    count ++;
                }
            L[first] = count - 1;
            R[count - 1] = first;
        }
        dancing(0);
        printf("%d
",ANS);
    }

    return    0;
}


void    ini(void)
{
    ANS = 0;
    R[HEAD] = 1;
    L[HEAD] = N;
    for(int i = 1;i <= N;i ++)
    {
        L[i] = i - 1;
        R[i] = i + 1;
        U[i] = D[i] = C[i] = i;
    }
    R[N] = 0;
}

void    dancing(int k)
{
    if(k + cut() <= ANS)
        return    ;
    if(R[HEAD] == HEAD)
    {
        ANS = ANS > k ? ANS : k;
        return    ;
    }

    int    c = R[HEAD];

    for(int i = D[c];i != c;i = D[i])
    {
        remove(i);
        for(int j = R[i];j != i;j = R[j])
            remove(j);
        dancing(k + 1);
        for(int j = L[i];j != i;j = L[j])
            resume(j);
        resume(i);
    }

    return    ;
}

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]] = i;
        R[L[i]] = i;
    }
}

void    debug(int count)
{
    for(int i = 0;i <= count;i ++)
        printf("U[%d]=%d D[%d]=%d L[%d]=%d R[%d]=%d c[%d]=%d
",i,U[i],i,D[i],i,L[i],i,R[i],i,C[i]);
    return    ;
}

int    cut(void)
{
    int    sum = 0;
    for(int i = R[HEAD];i;i = R[i])
        sum ++;
    return    sum;
}