Submission #1137880

#	Time	Username	Problem	Language	Result	Execution time	Memory
1137880		pan	Food Court (JOI21_foodcourt)	C++20	100 / 100	438 ms	58384 KiB

foodcourt

#include <bits/stdc++.h>
//#include "includeall.h"
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define endl '\n'
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define input(x) scanf("%lld", &x);
#define input2(x, y) scanf("%lld%lld", &x, &y);
#define input3(x, y, z) scanf("%lld%lld%lld", &x, &y, &z);
#define input4(x, y, z, a) scanf("%lld%lld%lld%lld", &x, &y, &z, &a);
#define print(x, y) printf("%lld%c", x, y);
#define show(x) cerr << #x << " is " << x << endl;
#define show2(x,y) cerr << #x << " is " << x << " " << #y << " is " << y << endl;
#define show3(x,y,z) cerr << #x << " is " << x << " " << #y << " is " << y << " " << #z << " is " << z << endl;
#define all(x) x.begin(), x.end()
#define discretize(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());
#define FOR(i, x, n) for (ll i =x; i<=n; ++i) 
#define RFOR(i, x, n) for (ll i =x; i>=n; --i) 
#pragma GCC optimize("O3","unroll-loops")
using namespace std;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
//using namespace __gnu_pbds;
//#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
//#define ordered_multiset tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>
typedef long long ll;
typedef long double ld;
typedef pair<ld, ll> pd;
typedef pair<string, ll> psl;
typedef pair<ll, ll> pi;
typedef pair<pi, ll> pii;
typedef pair<pi, pi> piii;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

struct query
{
	ll add, del, net;
	query(ll x = 0)
	{
		add = del = net = 0;
		if (x < 0) del -= x;
		else add += x;
	}
	query operator + (query const & other)
	{
		query ret;
		ll take = min(add - net, other.del - other.net);
		ret.net = net + other.net + take;
		ret.add = add + other.add;
		ret.del = del + other.del;
		return ret;
	}
};

struct node {
    int s, e, m; 
    query val; 
    node * l, * r; 
    node(int S, int E) { 
        s = S, e = E, m = (s + e) / 2;
        val = query();
        if (s != e) { 
            l = new node(s, m); 
            r = new node(m + 1, e); 
        }
    }
    void update(int X, ll V) { 
        if (s == e) val= query(V);
        else { 
            if (X <= m) l -> update(X, V); 
            else r -> update(X, V); 
            val = l->val + r->val;
        }
    }
    query get(int S, int E) {
        if (s == S && e == E) return val;
        else if (E <= m) return l -> get(S, E);
        else if (S >= m + 1) return r -> get(S, E);
        else return l -> get(S, m) + r -> get(m + 1, E);
    }
    ll bs(ll x, ll v = 0)
    {
		if (s==e) return (v + val.add >= x)?s:-1;
		if (v + l->val.add >= x) return l->bs(x, v);
		return r->bs(x, v + l->val.add);
	}
}* root;

ll n,m ,q, no = 0, no2 = 0;
vector<pi> upd1[250005], upd2[250005];
vector<pii> que[250005];
ll grp[250005], ans[250005];



int main()
{
	cin >> n >> m >> q;
	for (ll i=0; i<q; ++i) 
	{
		ll t;
		cin >> t;
		if (t==1)
		{
			ll l, r, c, k;
			cin >> l >> r >> c >> k;
			grp[++no] = c;
			upd1[l].pb(mp(k , no));
			upd1[r + 1].pb(mp(0,  no));
			
		}
		else if (t==2)
		{
			ll l, r, k;
			cin >> l >> r >> k;
			upd2[l].pb(mp(k, ++no));
			upd2[r + 1].pb(mp(0, no));
		}
		else
		{
			ll a, b;
			cin >> a >> b;
			que[a].pb(mp(mp(b, no), no2++));
		}
	}
	root = new node(0, no);
	for (ll i=1; i<=n; ++i)
	{
		for (pi u: upd1[i]) root->update(u.s, u.f);
		for (pi u: upd2[i]) root->update(u.s, -u.f);
		for (pii u: que[i])
		{
			//show(u.s);
			query res = root->get(0, u.f.s);
			ll idx = res.net;
			ll t = root->bs(idx + u.f.f);
			//show(u.f.f + idx);
			//show(t);
			if (t==-1 || t > u.f.s) ans[u.s] = 0;
			else ans[u.s] = grp[t];
		}
	}
	for (ll i=0; i<no2; ++i) cout << ans[i] << endl;
	return 0;
}

• Subtask 1: 2 / 2
• Subtask 2: 5 / 5
• Subtask 3: 7 / 7
• Subtask 4: 21 / 21
• Subtask 5: 15 / 15
• Subtask 6: 13 / 13
• Subtask 7: 26 / 26
• Subtask 8: 11 / 11