CodeForces 1140C 貪心+優先隊列

Playlist CodeForces - 1140C

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.

Input
4 3
4 7
15 1
3 6
6 8
Output
78

Input
5 3
12 31
112 4
100 100
13 55
55 50
Output
10000

思路：
貪心，盡力找最大的，因爲song有兩個屬性值，所以需要確定其中一個，來改變另一個。
先對輸入數據根據beauty值進行降序排序（這樣在向隊列中push數據時當前k個元素的最小beauty值即爲當前元素的beauty值），然後依次將song的length加入到優先隊列中，使用優先隊列來維護length即可，優先隊列設置爲升序排序，如果隊列中元素個數大於K，則pop（）出隊列頭部的元素，在將當前元素插入進去。
result = max ( result , sum_of_length * 當前元素的beauty值）

#include<iostream>
#include<vector>
#include<algorithm>
#include<queue>

using namespace std;


struct song{
    int t;
    int b;
};

bool cmp(song &a,song &bb){
    return a.b>bb.b;
}


int main(){
    int n,k;
    scanf("%d%d",&n,&k);
    vector<song> v;
    for(int i=0;i<n;i++){
        song s;
        scanf("%d%d",&s.t,&s.b);
        v.push_back(s);
    }
    sort(v.begin(),v.end(),cmp);


    priority_queue<int,vector<int>,greater<int>  >  q;

    long long len = 0;
    long long res = 0;
    for(int i=0;i<n;i++){

        if(q.size()>=k){
            len -= q.top();
            q.pop();
        }

        len += v[i].t;
        q.push(v[i].t);
        res = max(res,len*v[i].b);
    }

    printf("%lld\n",res);

    return 0;

}