# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
242607 | apostoldaniel854 | Schools (IZhO13_school) | C++14 | 171 ms | 9460 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define pb push_back
#define dbg(x) cerr << #x << " " << "\n"
/**
- we want to find to disjoint subsets of indices X (|X| = M) and Y (|Y| = S) so that sum{a[i]|i belongs to X} + sum{b[i]|i belongs to Y}
- it's easy to prove that if we sort the indices by b[i] - a[i] we will not use a a[i] and b[j] with i > j
- now we can see the best solution for a on prefix and the best solution for b on suffix and combine them
**/
const int N = 3e5;
int a[1 + N], b[1 + N];
int order[1 + N];
bool cmp (int x, int y) {
return b[x] - a[x] < b[y] - a[y];
}
ll pref[1 + N], suff[1 + N + 1];
int main () {
ios::sync_with_stdio (false);
cin.tie (0); cout.tie (0);
int n, M, S;
cin >> n >> M >> S;
for (int i = 1; i <= n; i++) {
cin >> a[i] >> b[i];
order[i] = i;
}
sort (order + 1, order + n + 1, cmp);
/// make pref
ll sum = 0;
priority_queue <int> pq;
if (M == 0)
pref[0] = 0;
else
pref[0] = -1;
for (int i = 1; i <= n; i++) {
int index = order[i];
pq.push (-a[i]);
sum += a[i];
if (pq.size () > M) {
sum += pq.top ();
pq.pop ();
}
if (pq.size () == M)
pref[i] = sum;
else
pref[i] = -1;
}
/// make suff
while (pq.size ()) pq.pop ();
sum = 0;
if (S == 0)
suff[n + 1] = 0;
else
suff[n + 1] = -1;
for (int i = n; i > 0; i--) {
int index = order[i];
pq.push (-b[i]);
sum += b[i];
if (pq.size () > S) {
sum += pq.top ();
pq.pop ();
}
if (pq.size () == S)
suff[i] = sum;
else
suff[i] = -1;
}
ll ans = 0;
for (int i = 0; i <= n; i++)
if (pref[i] != -1 && suff[i + 1] != -1)
ans = max (ans, pref[i] + suff[i + 1]);
cout << ans << "\n";
return 0;
}
Compilation message (stderr)
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