Codeforces Round #311 (Div. 2)C. Arthur and Table

C. Arthur and Table

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

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

枚举作为最大桌子腿的长度，比如第3个样例我们可以枚举 3 2 1。   当枚举到2 的时候，3就必然要删除...

所以我们可以 排序，multiset瞎搞了

/* ***********************************************
Author        :guanjun
Created Time  :2016/8/14 9:42:21
File Name     :cf311c.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 100010
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;
priority_queue<int,vector<int>,greater<int> >pq;
struct Node{
    int x,y;
};
struct cmp{
    bool operator()(int a,int b){
        return a>b;
    }
};
multiset<int,cmp>s;
vector<int>v[maxn];
int vis[maxn];
int x[maxn];
int y[maxn];
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    int n;
    cin>>n;
    int sum=0,Min=INF,Max=0;
    for(int i=1;i<=n;i++){
        scanf("%d",&x[i]);
    }
    for(int i=1;i<=n;i++){
        scanf("%d",&y[i]);
        v[x[i]].push_back(y[i]);
        sum+=y[i];
        s.insert(y[i]);
        Min=min(Min,x[i]);
        Max=max(Max,x[i]);
    }
    if(n==2){
        if(x[1]==x[2])cout<<0<<endl;
        else cout<<min(y[1],y[2])<<endl;return 0;
    }
    int ans=INF;
    for(int i=Max;i>=Min;i--){
        int m=v[i].size();
        if(m>0&&!vis[i]){
            vis[i]=1;
            int tmp=0,tmp2=0;
            for(int j=0;j<m;j++){
                auto pos=s.find(v[i][j]);
                s.erase(pos);
                tmp+=v[i][j];
            }
            int cnt=0;
            for(auto x:s){
                if(cnt==m-1)break;
                tmp+=x;
                cnt++;
            }
            ans=min(ans,sum-tmp);
        }
    }
    if(ans==INF){
        cout<<0<<endl;
    }
    else cout<<ans<<endl;
    return 0;
}

注意 multiset删除某个value时 他会把值为value的全部删除..

如果想只删除一个，那么可以加一个find，erase 一个值，具体看代码