이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
//#pragma GCC optimize("Ofast,unroll-loops,inline")
//#pragma GCC target("avx,avx2,sse3,ssse3,sse4.1,sse4.2,fma,bmi2,abm,popcnt,mmx,tune=native")
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define int ll
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
struct segment_tree {
vector<int> tree;
int n;
segment_tree(int n) : n(n) {
tree.resize(4 * n, -3e9);
}
int get(int v, int tl, int tr, int l, int r) {
if (tr < l || r < tl)
return -3e9;
if (l <= tl && tr <= r)
return tree[v];
int tm = (tl + tr) / 2;
return max(get(v * 2, tl, tm, l, r), get(v * 2 + 1, tm + 1, tr, l, r));
}
void upd(int v, int tl, int tr, int p, int x) {
if (tl == tr) {
tree[v] = x;
return;
}
int tm = (tl + tr) / 2;
if (tm >= p)
upd(v * 2, tl, tm, p, x);
else
upd(v * 2 + 1, tm + 1, tr, p, x);
tree[v] = max(tree[v * 2], tree[v * 2 + 1]);
}
int get(int l, int r) {
return get(1, 0, n - 1, l, r);
}
void upd(int p, int x) {
upd(1, 0, n - 1, p, x);
}
};
/*
* e_i e_j
* x_i x_j
* e_i - e_j >= x_j - x_i
* e_i + x_i >= x_j + e_j
*/
void solve() {
int n;
cin >> n;
vector<pair<int, int>> p(n);
for (int i = 0; i < n; ++i)
cin >> p[i].first >> p[i].second;
sort(all(p));
vector<array<int, 3>> q(n);
for (int i = 0; i < n; ++i)
q[i] = {p[i].second, i, p[i].first};
sort(rall(q));
// for (auto i : q)
// cout << i[0] << ' ';
// cout << '\n';
segment_tree neg(n), pos(n);
int res = 0;
for (int i = 0; i < n; ++i) {
// cout << "elem: " << q[i][2] << ' ' << q[i][0] << ' ' << q[i][1] << endl;
// cout << pos.get(0, q[i][1] - 1) << ' ' << neg.get(q[i][1] + 1, n - 1) << '\n';
if (pos.get(0, q[i][1] - 1) < q[i][0] + q[i][2] && neg.get(q[i][1] + 1, n - 1) < q[i][0] - q[i][2]) {
++res;
// cout << "now\n";
}
neg.upd(q[i][1], q[i][0] - q[i][2]);
pos.upd(q[i][1], q[i][0] + q[i][2]);
}
cout << res;
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
int t = 1;
// cin >> t;
while (t--)
solve();
}
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