Educational Codeforces Round 62 (Rated for Div. 2) C. Playlist（枚举+优先队列）

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 n
songs. The i-th song is characterized by two numbers ti and bi — 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 3 songs having lengths [5,7,4] and beauty values [11,14,6] is equal to (5+7+4)⋅6=96

.

You need to choose at most k

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 n
and k (1≤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 n
lines contains two integers ti and bi (1≤ti,bi≤106) — the length and beauty of i

-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 1,3,4
, so the total pleasure is (4+3+6)⋅6=78

.

In the second test case we can choose song 3
. The total pleasure will be equal to 100⋅100=10000

.

解析

按照美丽值排序，然后枚举歌曲，枚举到当前歌曲时，也就是在前面选取k-1个最长的歌曲，用优先队列处理一下就好了

code

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N=3e5+10;
ll ar[N];

struct node{
    ll a,b;
    friend bool operator <(node a,node b){
        return a.b>b.b||a.b==b.b&&a.a>b.a;
    }
}ap[N];

int main()
{
    //cout << "Hello world!" << endl;
    priority_queue<ll,vector<ll>,greater<ll> >pq;
    ll n,k,mx1=0;
    cin>>n>>k;
    for(int i=0;i<n;i++){
        cin>>ap[i].a>>ap[i].b;
        mx1=max(mx1,ap[i].a*ap[i].b);
    }
    sort(ap,ap+n);
    if(k==1){
        cout<<mx1<<endl;
        return 0;
    }
    ll cnt=1,now=ap[0].a*ap[0].b,mx=now,length=ap[0].a;
    pq.push(ap[0].a);
    for(int i=1;i<n;i++){
        now=ap[i].b*(length+ap[i].a);
        mx=max(mx,now);
        if(cnt<k-1){
            length+=ap[i].a;
            pq.push(ap[i].a);
            cnt++;
        }
        else {
            int t=pq.top();
            if(t<ap[i].a){
                pq.pop();
                pq.push(ap[i].a);
                length+=ap[i].a-t;
            }
        }

    }
    cout<<mx<<endl;

    return 0;
}