이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-O3")
#pragma GCC target("avx2")
using namespace std;
#define pb push_back
#define sz size()
#define all(x) x.begin(), x.end()
#define F first
#define S second
#define int long long
typedef pair < int, int > pii;
typedef vector < int > vi;
typedef vector < vi > vvi;
const int N = 200200;
int pref[N];
int t[4 * N];
void upd (int v, int l, int r, int pos, int val) {
if (l > pos || r < pos) {
return;
}
if (l == r) {
t[v] = val;
return;
}
int mid = (l + r) / 2;
upd (2 * v, l, mid, pos, val);
upd (2 * v + 1, mid + 1, r, pos, val);
t[v] = max(t[2 * v], t[2 * v + 1]);
}
int get_max (int v, int l, int r, int tl, int tr) {
if (l > tr || r < tl) {
return -1e18;
}
if (l >= tl && r <= tr) {
return t[v];
}
int mid = (l + r) / 2;
return max(get_max(2 * v, l, mid, tl, tr), get_max(2 * v + 1, mid + 1, r, tl, tr));
}
void solve(){
int n;
cin >> n;
pii a[n + 7];
for (int i = 1; i <= n; i++) {
cin >> a[i].F >> a[i].S;
}
sort (a + 1, a + n + 1);
for (int i = 1; i <= n; i++) {
pref[i] = pref[i - 1] + a[i].S;
upd (1, 1, n, i, pref[i] - a[i].F);
}
int ans = 0;
for (int i = 1; i <= n; i++){
int mx = get_max(1, 1, n, i, n);
ans = max(ans, mx - pref[i - 1] + a[i].F);
// cout << i << " " << mx << " " << ans << "\n";
}
cout << ans;
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
int t = 1;
// cin >> t;
while (t--) {
solve();
cout << "\n";
}
}
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