可除图的最大团 51Nod - 1524

​​https://www.51nod.com/Challenge/Problem.html#!#problemId=1524​​

题意可以转换为 用给定的几个数构成一个哈斯图 求最长链

类似素数筛地跑一遍 复杂度n*logn

下面两种写法 大同小异 都要加读入优化才能过

#include <bits/stdc++.h>
using namespace std;
const int maxn=1e6+10;

int prime[maxn],book[maxn],ary[maxn],cnt[maxn],maxx[maxn];
int num,n,lim,len;

template <class T>
inline void _cin(T &ret)
{
    char c;
    ret = 0;
    while((c = getchar()) < '0' || c > '9');
    while(c >= '0' && c <= '9')
    {
        ret = ret * 10 + (c - '0');
        c = getchar();
    }
}

void init()
{
    int i,j;
    book[1]=1;
    for(i=2;i<=lim;i++){
        if(!book[i]){
            for(j=i+i;j<=lim;j+=i){
                book[j]=1;
            }
            prime[++num]=i;
        }
    }
}

int main()
{
    int i,j,ans;
    //scanf("%d",&n);
    _cin(n);
    for(i=1;i<=n;i++){
        //scanf("%d",&ary[i]);
        _cin(ary[i]);
        lim=max(lim,ary[i]);
    }
    init();
    sort(ary+1,ary+n+1);
    ans=0;
    for(i=1;i<=n;i++){
        if(len==0||ary[len]!=ary[i]){
            ary[++len]=ary[i];
        }
        cnt[ary[len]]++;
        maxx[ary[len]]++;
        ans=max(ans,cnt[ary[len]]);
    }
    for(i=1;i<=lim;i++){
        for(j=1;j<=num&&i*prime[j]<=lim;j++){
            maxx[i*prime[j]]=max(maxx[i*prime[j]],maxx[i]+cnt[i*prime[j]]);
        }
    }
    for(i=1;i<=len;i++){
        ans=max(ans,maxx[ary[i]]);
    }
    printf("%d\n",ans);
    return 0;
}

#include <bits/stdc++.h>
using namespace std;
const int maxn=1e6+10;

int ary[maxn],cnt[maxn],maxx[maxn],book[maxn];
int n,len;

template <class T>
inline void _cin(T &ret)
{
    char c;
    ret = 0;
    while((c = getchar()) < '0' || c > '9');
    while(c >= '0' && c <= '9')
    {
        ret = ret * 10 + (c - '0');
        c = getchar();
    }
}

int main()
{
    int i,j,ans;
    //scanf("%d",&n);
    _cin(n);
    for(i=1;i<=n;i++){
        //scanf("%d",&ary[i]);
        _cin(ary[i]);
    }
    sort(ary+1,ary+n+1);
    ans=0;
    for(i=1;i<=n;i++){
        if(len==0||ary[len]!=ary[i]){
            ary[++len]=ary[i];
            book[ary[len]]=1;
        }
        cnt[ary[len]]++;
        maxx[ary[len]]++;
        ans=max(ans,cnt[ary[len]]);
    }
    for(i=1;i<=len;i++){
        for(j=2*ary[i];j<=1e6;j+=ary[i]){
            if(book[j]){
                maxx[j]=max(maxx[j],maxx[ary[i]]+cnt[j]);
                ans=max(ans,maxx[j]);
            }
        }
    }
    printf("%d\n",ans);

    return 0;
}