You can go back on the greedy

The first time this question, but I think it is not rigorous, hoping Sword friends feel the same about the exchange.

https://www.luogu.org/problem/P4053

I write code like this:

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define re register
inline long long read() {
    char ch = getchar(); long long x = 0, f = 1;
    while(ch < '0' || ch > '9') {
        if(ch == '-') f = -1;
        ch = getchar();
    } while('0' <= ch && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    } return x * f;
}
struct build {
    ll a;
    ll t;

} datas [150001];
inline bool cmp (const build &a,const build &b) {
    return a.t < b.t;
}
priority_queue <int> q;
signed main(){
#ifndef ONLINE_JUDGE
    freopen ("shit.txt","r",stdin);
#endif 
#ifdef ONLINE_JUDGE
#endif
    int n = read ();
    for (int i = 1;i <= n;++ i) {
        datas [i].a = read ();
        datas [i].t = read ();
    }
    sort (datas + 1,datas + 1 + n,cmp);
    //开始贪心
    int nowt = 0 ; 
    LL ANS = 0 , Nowa = 0 ;
     for ( int I = . 1 ; I <= n-; ++ I) {
         // actually BFS 
        IF (nowt DATAS + [I] II.A> DATAS [I] .T ) {
             // do not satisfy it comes out a repair and replacement longest 
            IF (q.empty (!) && q.top ()> DATAS [I] && nowt II.A - q.top () + DATAS [I] II.A <= DATAS [I] .T) { 
                nowt - = q.top (); 
                q.pop (); 
                q.push (DATAS [I] II.A); 
                nowt + = DATAS [I] II.A; 
            }
        } else {     // it is built directly 
            ++ ANS; 
            nowt + = DATAS [I] II.A; 
            q.push (DATAS [I] II.A); 
        } 
    } 
    COUT << ANS << endl;
     return  0 ; 
}

But I think it is very metaphysical. I could not say why 2333