Aimee

很水的题目

很简单的反悔贪心，能干就干，干不了就把最小的扔掉

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<queue>
#define int long long
using namespace std;
priority_queue <int> q;
int Aimee;
int n;
struct wor{
    int ti;
    int mo;
} th[100001];
int read(){
    int f=1,ans=0;
    char c;
    c=getchar();
    while(c!='-'&&(c<'0'||c>'9')){
        c=getchar();
    }
    if(c=='-'){
        f=-1;
        c='0';
    }
    while(c>='0'&&c<='9'){
        ans=(ans<<1)+(ans<<3)+c-'0';
        c=getchar();
    }
    return ans*f;
}
bool cmp( wor x,wor y){
    return x.ti<y.ti;
}
signed main(){
    n=read();
    for(int i=1;i<=n;++i){
        th[i].ti=read();
        th[i].mo=read(); 
    }
    sort(th+1,th+n+1,cmp);
    for(int i=1;i<=n;++i){
        q.push(-th[i].mo);
        Aimee+=th[i].mo;
        if(q.size()>th[i].ti){
            Aimee+=q.top();
            q.pop();
        }
    }
    printf("%lld",Aimee); 
    return 0;
}