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原创最新推荐文章于 2026-04-24 10:31:21 发布 791 阅读 AI 写同款· GEO 优化 ›

【Dify】PPT生成工具应用本文介绍了一款自动化PPT生成工具，通过智能接口将文本内容高效转化为结构化演示文档。该系统采用三阶段工作流（内容输入、HTTP请求上传、文件输出），无需复杂操作即可生成专业PPT。工具适用于教学课件、产品发布、企业培训等场景，能显著提升内容表达效率，特别适合初学者快速制作演示材料。文章详细说明了核心节点、应用场景及开发资源，为文本到PPT的自动化转换提供了实用解决方案。阅读详情

Optiver sponsored problem.

After years of hard work Optiver has developed a mathematical model that allows them to predict wether or not a company will be succesful. This obviously gives them a great advantage on the stock market.

In the past, Optiver made a deal with a big company, which forces them to buy shares of the company according to a fixed schedule. Unfortunately, Optiver's model has determined that the company will go bankrupt after exactly n days, after which their shares will become worthless.

Still, Optiver holds a large number of sell options that allows them to sell some of the shares before the company goes bankrupt. However, there is a limit on the number of shares Optiver can sell every day, and price Optiver receives for a share may vary from day to day. Therefore, it is not immediately clear when Optiver should sell their shares to maximize their profit, so they asked you to write a program to calculcate this.

Input

On the first line an integer t (1 <= t <= 100): the number of test cases. Then for each test case:

One line with an integer n (1 <= n <= 100 000): the number of days before the company goes bankrupt.

n lines with three integers x_i (0 <= x_i <= 100), p_i (0 <= p_i <= 100) and m_i (0 <= m_i <= 10 000 000): the number of shares Optiver receives on day i, the (selling) price per share on day i, and the maximum number of shares Optiver can sell on day i, respectively.

Output

For each test case:

One line with the maximum profit Optiver can achieve.

Sample Input

1
6
4 4 2
2 9 3
2 6 3
2 5 9
2 2 2
2 3 3

Sample Output

题意：有n张股票，给出每天股票的买进数量，当天的股票价格和当天最大抛出量，第i天得到的股票当天可以不抛，可以留到以后抛。问这n天最多能卖多少钱？

思路：贪心，从后往前贪心，最后一天的股票当然只能在最后一天卖出，第i天的可以在第i天及以后卖出，那么就可以维护一个优先队列来存放第i天及以后的天数中抛出量不为零的日期（价格高的优先），那抛出第i天时先从优先队列中取出价格最高的。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>


#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
//typedef __int64 ll;

#define fre(i,a,b)  for(i = a; i <b; i++)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define bug         pf("Hi\n")

using namespace std;

#define INF 0x3f3f3f3f
#define N 100005

int r[N],p[N],s[N];
int n;


struct stud{
 int p,num;
 friend  bool operator < (stud a,stud b)
    {
    	return a.p<b.p;
	}
};

priority_queue<stud>q;


void solve()
{
	int i,j;
	while(!q.empty()) q.pop();

	int ans=0;

	struct stud cur;

	for(i=n-1;i>=0;i--)
	{
        cur.num=s[i];
        cur.p=p[i];
        q.push(cur);

		while(!q.empty()&&r[i])
		{
		   	 cur=q.top();

		   	 q.pop();

		   	 if(cur.num>r[i])
			 {
			 	ans+=cur.p*r[i];
			 	//pf("%d %d\n",cur.p,r[i]);
			 	cur.num-=r[i];
			 	r[i]=0;
			 	q.push(cur);
			 }
             else
			 {
			 	ans+=cur.p*cur.num;
			 	r[i]-=cur.num;
			 //	pf("%d %d\n",cur.p,cur.num);

			 }
		}

	}

	pf("%d\n",ans);
}
int main()
{
	int i,j,t;
	sf(t);
	while(t--)
	{
		sf(n);
		fre(i,0,n)
		 sfff(r[i],p[i],s[i]);
        solve();
	}
   return 0;

}