제출 #946344

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
946344		vqpahmad	Examination (JOI19_examination)	C++14	0 / 100	192 ms	16324 KiB

examination

#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#ifdef ONPC
#include"debug.h"
#else
#define debug(...) 42
#endif
#define endl '\n'
#define ll long long
#define pii pair<int,int>
#define F first
#define S second
#define pb push_back
#define sz(a) (int)a.size()
#define all(a) a.begin(),a.end()
template<class T> bool ckmin(T& a, const T& b) { return b < a ? a = b, 1 : 0; }
template<class T> bool ckmax(T& a, const T& b) { return a < b ? a = b, 1 : 0; }
const int mod = 1e9 + 7;
const int MAXN = 1e6 + 15;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
int ans[MAXN];
template<class T> using order_set = tree<T, null_type, less<>, rb_tree_tag, tree_order_statistics_node_update>;
struct point {
	int x, y;
};
int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n, q;
	cin >> n >> q;
	vector<point> a(n);
	vector<ll> b(n);
	for (int i = 0; i < n; i++){
		cin >> a[i].x >> a[i].y;
		b[i] = a[i].x + a[i].y;
	}
	sort(all(b));
	vector<array<int, 3>> queries(q);
	vector<array<int, 3>> queriesA(q);
	vector<array<int, 3>> queriesB(q);
	for (int i = 0; i < q; i++){
		cin >> queries[i][0] >> queries[i][1] >> queries[i][2];
		queries[i][2] = max(queries[i][2], queries[i][0] + queries[i][1]);
		queriesA[i] = {queries[i][2], queries[i][0], i};
		queriesB[i] = {queries[i][2], queries[i][1], i};
		ans[i] += lower_bound(all(b), queries[i][2]) - b.begin();
	}
	sort(all(queriesA), greater<>());
	sort(all(queriesB), greater<>());
	sort(all(a), [](point lh, point rh){
			return lh.x + lh.y > rh.x + rh.y;
			});
	int p = 0;
	order_set<pii> st;
	for (int i = 0; i < q; i++){// solve for b
		while (p < n && a[p].x + a[p].y > queriesA[i][0]){
			st.insert({a[p].x, p});
			p++;
		}
		ans[queriesA[i][2]] += st.order_of_key({queriesA[i][1], -1});
	}
	st.clear();
	p = 0;
	for (int i = 0; i < q; i++){// solve for b
		while (p < n && a[p].x + a[p].y > queriesB[i][0]){
			st.insert({a[p].y, p});
			p++;
		}
		ans[queriesB[i][2]] += st.order_of_key({queriesB[i][1], -1});
	}
	for (int i = 0; i < q; i++){
		cout << n - ans[i] << endl;
	}
}

• Subtask 1: 0 / 2
• Subtask 2: 0 / 20
• Subtask 3: 0 / 21
• Subtask 4: 0 / 57