1 solutions

  • 2
    @ 2026-4-26 11:18:22
    #include<bits/stdc++.h>
    using namespace std;
    const int N=1e5+10;
    typedef pair<int,int> PII;
    #define g first
    #define d second
    PII cow[N];
    int main()
    {
    	int n;
    	cin>>n;
    	for(int i=1;i<=n;i++)
    	{
    		cin>>cow[i].g>>cow[i].d;
    	}
    	sort(cow+1,cow+n+1,[](PII a,PII b){
    		return a.d<b.d;
    	});
    	priority_queue<int,vector<int>,greater<int>> q;
    	int total=0;
    	for(int i=1;i<=n;i++)
    	{
    		if(q.size()<cow[i].d) //可以直接轩 
    		{
    			q.push(cow[i].g);
    			total+=cow[i].g;
    		}
    		else if(q.size()&&q.top()<cow[i].g)
    		{
    			//时间已满,但是当前奶牛的产量比最小的小
    			total-=q.top();
    			q.pop();//返回
    			q.push(cow[i].g); //选择新的奶牛 
    			total+=cow[i].g; 
    		}
    	}
    	cout<<total;
    	return 0; 
    }
    
    • 1

    Information

    ID
    3055
    Time
    1000ms
    Memory
    256MiB
    Difficulty
    5
    Tags
    (None)
    # Submissions
    11
    Accepted
    4
    Uploaded By