CF 337E(Divisor Tree-枚举树节点的父亲)

最新推荐文章于 2021-12-04 10:38:47 发布

原创最新推荐文章于 2021-12-04 10:38:47 发布 · 2k 阅读

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阶乘算法

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本文介绍了一种算法，用于构建包含特定整数集合的最小分拆树。该算法通过枚举树结构来寻找最优解，并确保每个输入整数至少出现在树的一个节点上。文章还提供了实现这一算法的具体代码。

A divisor tree is a rooted tree that meets the following conditions:

Each vertex of the tree contains a positive integer number.
The numbers written in the leaves of the tree are prime numbers.
For any inner vertex, the number within it is equal to the product of the numbers written in its children.

Manao has n distinct integers a₁, a₂, ..., a_n. He tries to build a divisor tree which contains each of these numbers. That is, for each a_i, there should be at least one vertex in the tree which contains a_i. Manao loves compact style, but his trees are too large. Help Manao determine the minimum possible number of vertices in the divisor tree sought.

Input

The first line contains a single integer n (1 ≤ n ≤ 8). The second line contains n distinct space-separated integers a_i (2 ≤ a_i ≤ 10¹²).

Output

Print a single integer — the minimum number of vertices in the divisor tree that contains each of the numbers a_i.

Sample test(s)

input
2
6 10

output
7

input
4
6 72 8 4

output
12

input
1
7

output
1

Note

Sample 1. The smallest divisor tree looks this way:

Sample 2. In this case you can build the following divisor tree:

Sample 3. Note that the tree can consist of a single vertex.

NOI就因为这个少拿10分了啊。。。现在又因为这个掉Rating了吖囧。。。。

首先明确一点——O(n!) 在n≤10时可过，O(2^n)在n≤20时可过。。。

然后枚举树（都懂得。。）

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (8+10)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
int n,fa[MAXN]={0},p[MAXN]={0};
ll a[MAXN],a2[MAXN],ans=INF;
void dfs(int l)
{
	if (l==n+1)
	{
	//	For(i,n) cout<<a2[i]<<' ';cout<<endl;
	//	For(i,n) cout<<fa[i]<<' ';cout<<endl;
		
		ll tot=0;
		For(i,n) if (!((!p[i])&&fa[i])) tot++; 
		bool b=0;
		For(i,n-1) if (!fa[i]) {b=1;break;}
		tot+=b;
		For(i,n) tot+=p[i]*!(bool)fa[i];
		ans=min(ans,tot);
		
	//	cout<<tot<<endl;
		/*
		if (tot==9) 
		{
			cout<<'d';
		}*/
		return;
	}
	fa[l]=0;dfs(l+1);
	Fork(i,l+1,n) if (a2[i]%a[l]==0) fa[l]=i,a2[i]/=a[l],dfs(l+1),a2[i]*=a[l];	
}
int main()
{
//	freopen("Divisor.in","r",stdin);
	cin>>n;
	For(i,n) cin>>a[i];
	sort(a+1,a+1+n);
	For(i,n) 
	{
		ll x=a[i];
		for(ll j=2;j*j<=x;j++)
			while (x%j==0) p[i]++,x/=j;
		if (x>1) p[i]++,x=1;
		if (p[i]==1) p[i]--;	
	}
	memcpy(a2,a,sizeof(a));
	dfs(1);cout<<ans<<endl;
	
	return 0;
}