CF 557C(Arthur and Table-直接暴力)

最新推荐文章于 2020-04-04 20:36:07 发布

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本文探讨了如何通过移除部分腿使餐桌稳定的问题，并提供了一个算法解决方案。考虑到餐桌腿的数量和长度，以及移除每条腿所需的能量，算法的目标是最小化总能量消耗。

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Arthur has bought a beautiful big table into his new flat. When he came home, Arthur noticed that the new table is unstable.

In total the table Arthur bought has n legs, the length of the i-th leg is li.

Arthur decided to make the table stable and remove some legs. For each of them Arthur determined number di — the amount of energy that he spends to remove the i-th leg.

A table with k legs is assumed to be stable if there are more than half legs of the maximum length. For example, to make a table with 5 legs stable, you need to make sure it has at least three (out of these five) legs of the maximum length. Also, a table with one leg is always stable and a table with two legs is stable if and only if they have the same lengths.

Your task is to help Arthur and count the minimum number of energy units Arthur should spend on making the table stable.

Input

The first line of the input contains integer n (1 ≤ n ≤ 105) — the initial number of legs in the table Arthur bought.

The second line of the input contains a sequence of n integers li (1 ≤ li ≤ 105), where li is equal to the length of the i-th leg of the table.

The third line of the input contains a sequence of n integers di (1 ≤ di ≤ 200), where di is the number of energy units that Arthur spends on removing the i-th leg off the table.

Output

Print a single integer — the minimum number of energy units that Arthur needs to spend in order to make the table stable.

Sample test(s)

input
2
1 5
3 2

output
2

input
3
2 4 4
1 1 1

output
0

input
6
2 2 1 1 3 3
4 3 5 5 2 1

output
8

直接暴力找就行。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<functional>
#include<algorithm>
#include<iostream>
using namespace std;
#define MAXN (100000+10)
#define INF (2139062143)
#define For(i,n) for(int i=1;i<=n;i++)
int n,l[MAXN],d[MAXN];
pair<int,int> a[MAXN],b[MAXN];

int main()
{
//  freopen("CF557C.in","r",stdin);
    cin>>n;
    int tot=0;
    For(i,n) scanf("%d",&l[i]);
    For(i,n) scanf("%d",&d[i]),tot+=d[i];
    a[0]=b[0]=make_pair(0,0);
    a[n+1]=b[n+1]=make_pair(INF,INF);
    For(i,n) a[i]=make_pair(d[i],l[i]);
    For(i,n) b[i]=make_pair(l[i],d[i]);

    sort(a+1,a+1+n);
    sort(b+1,b+1+n);

    int p=0,len=0,ans=0;
    For(i,n)
    {
        p+=b[i].second;len++;
        if (b[i].first!=b[i+1].first)
        {
            int j=n;
            while (j&&len>1)
            {
                if (a[j].second<b[i].first) p+=a[j].first,len--;
                j--;
            }
            ans=max(ans,p);
            p=len=0;
        }
    }
    cout<<tot-ans<<endl;
    return 0;
}