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 thei-th leg isl_i.

Arthur decided to make the table stable and remove some legs. For each of them Arthur determined numberd_i — the amount of energy that he spends to remove thei-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 with5 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 integersl_i (1 ≤ l_i ≤ 10⁵), wherel_i is equal to the length of thei-th leg of the table.

The third line of the input contains a sequence of n integersd_i (1 ≤ d_i ≤ 200), whered_i is the number of energy units that Arthur spends on removing thei-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

21 53 2

Output

2

Input

32 4 41 1 1

Output

0

Input

62 2 1 1 3 34 3 5 5 2 1

Output
8

题意：给你一张桌子，有n条腿，告诉每条腿的长度（l）以及砍掉这条腿的花费（d），当最长腿的数量超过总数量的1/2，那么合法。问如何使得花费最小达到合法情况。
分析：暴力枚举。假设以腿长为a作为最长腿且腿a共有y条，计算花费cost需分两步：
（1）、腿长大于a的，都要砍掉，则cost+=bigger[a]；（bigger[a]可预处理得到）
（2）、腿长小于a的，假设有x条，则需要从小到大依次减去x-y+1条。由于花费d的范围只有1~200，所以可以利用花费来计算。
题目链接：http://codeforces.com/contest/557/problem/C
代码清单：

#include<queue>#include<stack>#include<cstdio>#include<iostream>#include<algorithm>using namespace std;typedef long long ll;const int maxn = 1e5 + 5;const ll maxv = 1e10 + 5;int n;struct Edge{int l,d;}leg[maxn];int cost[maxn]; ll Back[maxn];int num[maxn]; bool cmp(Edge a,Edge b){    return a.l<b.l;}int main(){    scanf("%d",&n);    for(int i=0;i<n;i++){        scanf("%d",&leg[i].l);        num[leg[i].l]++;    }    for(int i=0;i<n;i++)        scanf("%d",&leg[i].d);    sort(leg,leg+n,cmp);    ll sum=0;    for(int i=n-1;i>0;i--){ //计算腿长比leg[i-1].l大的总花费        sum+=(ll)leg[i].d;        if(leg[i].l!=leg[i-1].l){            Back[leg[i-1].l]=sum;        }    }    ll ans=Back[leg[0].l];    cost[leg[0].d]++;    for(int i=1;i<n;i++){        if(leg[i].l!=leg[i-1].l){            int an=i-num[leg[i].l]+1;            sum=Back[leg[i].l];            if(an>0){                for(int j=1;j<=200;j++){ //从小到大依次减去an数量的腿                    if(cost[j]){                        if(cost[j]>=an){                            sum+=(an*j);                            an=0;                            break;                        }                        else{                            sum+=(cost[j]*j);                            an-=cost[j];                        }                    }                }            }            ans=min(ans,sum);        }        cost[leg[i].d]++;    }    printf("%I64d\n",ans);    return 0;}