CodeForces - 566F Clique in the Divisibility Graph
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As you must know, the maximum clique problem in an arbitrary graph is NP-hard. Nevertheless, for some graphs of specific kinds it can be solved effectively.
Just in case, let us remind you that a clique in a non-directed graph is a subset of the vertices of a graph, such that any two vertices of this subset are connected by an edge. In particular, an empty set of vertexes and a set consisting of a single vertex, are cliques.
Let‘s define a divisibility graph for a set of positive integers A?=?{a1,?a2,?...,?an}as follows. The vertices of the given graph are numbers from set A, and two numbers ai and aj (i?≠?j) are connected by an edge if and only if either ai is divisible by aj, or aj is divisible by ai.
You are given a set of non-negative integers A. Determine the size of a maximum clique in a divisibility graph for set A.
Input
The first line contains integer n (1?≤?n?≤?106), that sets the size of set A.
The second line contains n distinct positive integers a1,?a2,?...,?an (1?≤?ai?≤?106) — elements of subset A. The numbers in the line follow in the ascending order.
Output
Print a single number — the maximum size of a clique in a divisibility graph for set A.
Examples
8
3 4 6 8 10 18 21 24
3
Note
In the first sample test a clique of size 3 is, for example, a subset of vertexes {3,?6,?18}. A clique of a larger size doesn‘t exist in this graph.
我们不妨把边看成有向的,当 a[i]|a[j] 时从 i 向 j 连边(要先把重复元素缩成一个点),这样我们发现随便一条链上的元素 都可以成为一个团,并且这是一个DAG。
所以我们直接找最长链好了。
#include<bits/stdc++.h> #define ll long long using namespace std; const int maxn=1000000; int n,cnt[maxn+5],f[maxn+5]; inline int read(){ int x=0; char ch=getchar(); for(;!isdigit(ch);ch=getchar()); for(;isdigit(ch);ch=getchar()) x=x*10+ch-‘0‘; return x; } int main(){ n=read(); for(int i=1;i<=n;i++) cnt[read()]++; for(int i=maxn;i;i--){ for(int j=i;j<=maxn;j+=i) f[i]=max(f[i],f[j]); f[i]+=cnt[i]; } printf("%d\n",f[1]); return 0; }
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