题目链接:http://codeforces.com/contest/1315/problem/D
思路:对于每一个数如果重复出现肯定是选花费最大的先增加,所以这几乎就是模拟了,只是暴力模拟的话,复杂度明显会炸掉,需要用可并堆优化一下,然而我不想写可并堆,所以就上pbds了,并查集好像也能做,但是当时没脑子
#pragma GCC optimize(2)
#include <bits/stdc++.h>
#include <ext/pb_ds/priority_queue.hpp>
using namespace __gnu_pbds;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll inff = 0x3f3f3f3f3f3f3f3f;
#define FOR(i,a,b) for(int i(a);i<=(b);++i)
#define FOL(i,a,b) for(int i(a);i>=(b);--i)
#define REW(a,b) memset(a,b,sizeof(a))
#define inf int(0x3f3f3f3f)
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%I64d",&a)
#define sd(a) scanf("%lf",&a)
#define ss(a) scanf("%s",a)
#define mod ll(1e9+7)
#define pb push_back
#define eps 1e-6
#define lc d<<1
#define rc d<<1|1
#define Pll pair<ll,ll>
#define P pair<int,int>
#define pi acos(-1)
#include <iostream>
using namespace std;
const int N=2e5+8;
ll n,b[N],ans,num[N];
Pll a[N];
unordered_map<ll,ll>mp;
__gnu_pbds::priority_queue<ll>g[N];
int main()
{
cin.tie(0);
cout.tie(0);
cin>>n;
FOR(i,1,n) sl(a[i].first),b[i]=a[i].first;
FOR(i,1,n) sl(a[i].second);
sort(b+1,b+n+1);
int qw=unique(b+1,b+n+1)-b;
b[qw]=inff;
FOR(i,1,qw) mp[b[i]]=i;
FOR(i,1,n) g[mp[a[i].first]].push(a[i].second),num[mp[a[i].first]]+=a[i].second;
FOR(i,1,qw-1)
{
int qw=mp[b[i]];
ll x=b[i+1]-b[i],y=0;
while(!g[qw].empty()&&x) ans+=y*g[qw].top(),num[qw]-=g[qw].top(),g[qw].pop(),x--,y++;
g[mp[b[i+1]]].join(g[qw]);
ans+=num[qw]*y;
num[mp[b[i+1]]]+=num[qw];
}
cout<<ans<<endl;
return 0;
}