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.
这个题目要好好想想，一开始以为是dp，因为数据太大可以发现肯定不是dp，后来想到用贪心，但是并不会写。。。
实际上这个是贪心+优先队列，用优先队列来处理，然后进行贪心。
优先队列怎么处理呢，首先我们对b进行排序，b大的放在前面，然后把每一个数的t值都放进队列里面，t越小放在越前面，
然后如果这个队列的size大于k，就要清除头部的那个元素，但是这个时候之后的取最大值并不会受到影响，因为如果是刚刚放进去
的被pop出来了，那sum肯定没有变大，而且同时b变小了，所以这个时候就不会对这里进行转移，如果是别的元素pop出来了，
那就没有影响了。
这里要想清楚为什么是对b进行排序，想清楚为什么对b进行排序之后进行优先队列的处理，就是最优的。
这个我之前没有想清楚，所以就写题目的时候就有各种天马行空的想法。。。
#include<iostream>
#include<cstdio>
#include<cstring>
#include <vector>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long ll;
const int maxn = 3e5 + 100;
struct node
{
int t, b;
node(int t=0,int b=0):t(t),b(b){}
bool operator<(const node&a)const
{
if (a.b == this->b) return a.t < this->t;
return a.b < this->b;
}
}exa[maxn];
priority_queue<ll, vector<ll>, greater<ll>>que;
int main()
{
ll sum = 0,ans=0;
int n, k;
cin >> n >> k;
for (int i = 1; i <= n; i++) scanf("%d%d", &exa[i].t, &exa[i].b);
sort(exa + 1, exa + 1 + n);
for(int i=1;i<=n;i++)
{
sum += exa[i].t;
que.push(exa[i].t);
if(que.size()>k)
{
sum -= que.top();
que.pop();
}
//printf("exa[%d]=%d %lld %lld\n", i, exa[i].t, sum,ans);
ans = max(ans, sum*exa[i].b);
}
printf("%lld\n", ans);
return 0;
}
原文地址：https://www.cnblogs.com/EchoZQN/p/10630161.html