C. Playlist
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You have a playlist consisting of nn songs. The ii-th song is characterized by two numbers titi and bibi — its length and beauty respectively. The pleasure of listening to set of songs is equal to the total length of the songs in the set multiplied by the minimum beauty among them. For example, the pleasure of listening to a set of 33 songs having lengths [5,7,4][5,7,4] and beauty values [11,14,6][11,14,6] is equal to (5+7+4)⋅6=96(5+7+4)⋅6=96.
You need to choose at most kk songs from your playlist, so the pleasure of listening to the set of these songs them is maximum possible.
Input
The first line contains two integers nn and kk (1≤k≤n≤3⋅1051≤k≤n≤3⋅105) – the number of songs in the playlist and the maximum number of songs you can choose, respectively.
Each of the next nn lines contains two integers titi and bibi (1≤ti,bi≤1061≤ti,bi≤106) — the length and beauty of ii-th song.
Output
Print one integer — the maximum pleasure you can get.
Examples
input
Copy
4 3 4 7 15 1 3 6 6 8
output
Copy
78
input
Copy
5 3 12 31 112 4 100 100 13 55 55 50
output
Copy
10000
Note
In the first test case we can choose songs 1,3,41,3,4, so the total pleasure is (4+3+6)⋅6=78(4+3+6)⋅6=78.
In the second test case we can choose song 33. The total pleasure will be equal to 100⋅100=10000100⋅100=10000.
题意：每首歌有两种属性：长度、美度。
给你n首歌，让你选k首使得舒适度最大；
舒适度：如有k首歌 舒适度 = k首歌的总长度*k首歌中最小的美度；
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 1e6 + 7;
//ll x[N], y[N];
struct Node {
ll x, y;
bool operator<(const Node &a) const {
if (y == a.y) return x > a.x;
return y > a.y;
}
} p[N];
priority_queue<ll, vector<ll>, greater<int> > q; //小的在前！！！
int main() {
int n, k;
scanf ("%d %d", &n, &k);
for(int i = 1; i <= n; ++i) {
scanf ("%I64d %I64d", &p[i].x, &p[i].y);
}
sort(p+1, p+n+1); //美度大的在前
ll sum = 0;
ll ans = 0;
for (int i = 1; i <= n; ++i) {
sum += p[i].x;
q.push(p[i].x);
if (q.size() > k) {
sum -= q.top();
//printf ("debug: %lld\n", q.top());
q.pop();
}
ans = max(ans, p[i].y*sum);
}
printf ("%I64d\n", ans);
return 0;
}