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>
using namespace std;
#define pb push_back
#define mp make_pair
#define f first
#define s second
#define ll long long
#define ld long double
#define all(_v) _v.begin(), _v.end()
#define sz(_v) (int)_v.size()
#define pii pair <int, int>
#define pll pair <ll, ll>
#define veci vector <int>
#define vecll vector <ll>
const int dx[4] = {1, -1, 0, 0};
const int dy[4] = {0, 0, -1, 1};
const double PI = 3.1415926535897932384626433832795;
const double eps = 1e-9;
const int MOD1 = 1e9 + 7;
const int MOD2 = 998244353;
const int MAXN = 2e5 + 10;
int n;
ll a[MAXN], b[MAXN];
int ord[MAXN];
ll pref[MAXN];
bool cmp(int i, int j) {
if(a[i] != a[j]) return (a[i] < a[j]);
return (b[i] < b[j]);
}
void solve() {
cin >> n;
for(int i = 1; i <= n; ++i)
cin >> a[i] >> b[i], ord[i] = i;
sort(ord + 1, ord + n + 1, &cmp);
for(int i = 1; i <= n; ++i)
pref[i] = pref[i - 1] + b[ord[i]];
ll ans = -1e18, res = -a[ord[1]];
/**
MAX(p[r] - p[l - 1] - a[r] + a[l])
MAX(p[r] - a[r] - (p[l - 1] - a[l]))
*/
for(int i = 1; i <= n; ++i) {
ans = max(ans, pref[i] - a[ord[i]] - res);
res = min(res, pref[i - 1] - a[ord[i]]);
}
cout << ans;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int CNT_TESTS = 1;
///cin >> CNT_TESTS;
for(int NUMCASE = 1; NUMCASE <= CNT_TESTS; ++NUMCASE) {
solve();
if(NUMCASE != CNT_TESTS) cout << '\n';
}
return 0;
}
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