Submission #1252113

#	Time	Username	Problem	Language	Result	Execution time	Memory
1252113		al95ireyiz	Fountain (eJOI20_fountain)	C++20	30 / 100	142 ms	29832 KiB

fountain

//*** Bismillah ***//
#pragma GCC optimize("O3", "fast-math", "unroll-loops", "no-stack-protector")
#include <bits/stdc++.h>
using namespace std;
#if !defined(ONLINE_JUDGE) and !defined(EVAL)
#include "template/debug.h"
#else
#define d(x...)
#endif
#define fr first
#define er erase
#define sc second
#define in insert
#define ll long long
#define pb push_back
#define vll vector<ll>
#define pll pair<ll,ll>
#define ull unsigned ll
#define vpll vector<pll>
#define len(x) (ll) x.size()
#define all(x) x.begin(),x.end()
const ll INF = 1e9;
const ll INFL = 1e18;
const ll MOD = 1e9+7;
// const ll MOD = 998244353;
const ll maxn = 1e5+5;
ll n,m,k=0;
ll a[maxn], b[maxn], to[20][maxn], up[20][maxn];
void _(ll &tt){
    cin >> n >> m;
    for(ll i = 1; i <= n; i ++){
        cin >> a[i] >> b[i];
    }
    memset(to, -1, sizeof(to));
    stack<pll>st;
    for(ll i = 1; i <= n; i ++){
        while(!st.empty() and st.top().fr <= a[i]){
            to[0][st.top().sc] = i;
            st.pop();
        }
        st.push({a[i], i});
        up[0][i] = b[i];
    }
    while(!st.empty()){
        to[0][st.top().sc] = 0;
        st.pop();
    }
    for(ll i = 1; i < 20; i ++){
        for(ll j = 0; j <= n; j ++){
            if(to[i - 1][j] == -1) continue;
            if(to[i - 1][to[i - 1][j]] == -1) continue;
            to[i][j] = to[i - 1][to[i - 1][j]];
            up[i][j] = up[i - 1][j] + up[i - 1][to[i - 1][j]];
        }
    }
    for(ll i = 1, a, b; i <= m; i ++){
        cin >> a >> b;
        ll cv = a;
        for(ll i = 19; i >= 0; i --){
            if(to[i][a] == -1) continue;
            if(up[i][a] < b){
                b -= up[i][a];
                a = to[i][a];
            }
        }
        cout << a << '\n';
    }
}
signed main() {
    ll tm = clock();
    cin.tie(0)->sync_with_stdio(0);
    ll t = 1;
    // cin >> t;
    for(ll tt = 1; tt <= t; tt ++) {
        _(tt);
    }
    cerr << "\n\033[1;31mTime: \033[1;30m" \
         << (double)(clock()-tm)/1000000 << "\033[1;32m seconds\n";
}

• Subtask 1: 0 / 30
• Subtask 2: 30 / 30
• Subtask 3: 0 / 40