HDU-6438-Buy and Resell（贪心，物品买卖）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6438

Problem Description

The Power Cube is used as a stash of Exotic Power. There are n cities numbered 1,2,…,n where allowed to trade it. The trading price of the Power Cube in the i -th city is ai dollars per cube. Noswal is a foxy businessman and wants to quietly make a fortune by buying and reselling Power Cubes. To avoid being discovered by the police, Noswal will go to the i -th city and choose exactly one of the following three options on the i -th day:

1. spend ai dollars to buy a Power Cube
2. resell a Power Cube and get ai dollars if he has at least one Power Cube
3. do nothing

Obviously, Noswal can own more than one Power Cubes at the same time. After going to the n cities, he will go back home and stay away from the cops. He wants to know the maximum profit he can earn. In the meanwhile, to lower the risks, he wants to minimize the times of trading (include buy and sell) to get the maximum profit. Noswal is a foxy and successful businessman so you can assume that he has infinity money at the beginning.

Input

There are multiple test cases. The first line of input contains a positive integer T (T≤250 ), indicating the number of test cases. For each test case:
The first line has an integer n . (1≤n≤105 )
The second line has n integers a1,a2,…,an where ai means the trading price (buy or sell) of the Power Cube in the i -th city. (1≤ai≤109 )
It is guaranteed that the sum of all n is no more than 5×105 .

Output

For each case, print one line with two integers —— the maximum profit and the minimum times of trading to get the maximum profit.

Sample Input

1 2 10 9

9 5 9 10 5

2 1

Sample Output

16 4

5 2

0 0

Hint

In the first case, he will buy in 1, 2 and resell in 3, 4. profit = - 1 - 2 + 10 + 9 = 16 In the second case, he will buy in 2 and resell in 4. profit = - 5 + 10 = 5 In the third case, he will do nothing and earn nothing. profit = 0

题目大意：一个常见的物品买卖问题，（话说最近碰见的物品买卖好多啊，要不要总结一个专题？？）

一个人，开局无限多的金钱，只能往前走，不能后退，他碰到的商品可以选择买或者买，最后要求能够获得的收益最大，同时买卖的次数最少；

牛客网2018多校练习赛有一个这样的题，也是商品买卖，可以看一看；手中只能拿一个的物品买卖

还有CF的一道题（这个是我看别人的链接，就顺便写了写，就当专项训练了）：一次只能交易一次的物品买卖

思路都差不多：每输入一个数，先买入，然后判断前面有没有比他小的数，有的话，前面的那个卖出，这个再次买入，相当于前面的那个变成了这个，然后标记一下为中间变量，如果后面的能用到，就相当于没有它，如果后面的用不到，则相当于它就是最后的结果，不用管；

因为我们是用的加减法算最大收益之和，因此1,2,10和1,10的收益是相同的，只考虑买卖次数即可；

对于次数，碰见中间变量减去1即可；

ac：

#include<stdio.h>
#include<string.h>  
#include<math.h>  
  
#include<map>   
//#include<set>
#include<deque>  
#include<queue>  
#include<stack>  
#include<bitset> 
#include<string>  
#include<fstream>
#include<iostream>  
#include<algorithm>  
using namespace std;  

#define ll long long  
#define INF 0x3f3f3f3f  
//#define mod 1e9+7
//#define max(a,b) (a)>(b)?(a):(b)
//#define min(a,b) (a)<(b)?(a):(b) 
#define clean(a,b) memset(a,b,sizeof(a))// 水印 
//std::ios::sync_with_stdio(false);

int n;

struct cmp{
	bool operator ()(const ll &a,const ll &b){
		return a>b;
	}
};

int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		priority_queue<ll,vector<ll>,cmp> que;
		map<ll,int> vis;//ll价格做中间商品的数量int 
		vis.clear();
		while(que.size())
			que.pop();
		ll ans=0,cnt=0;
		scanf("%d",&n);
		for(int i=1;i<=n;++i)
		{
			ll x;
			scanf("%lld",&x);
			que.push(x);
			if(que.top()<x)//可以交易 
			{
				cnt++;//交易 
				ans=ans+x-que.top();//收入 
				if(vis[que.top()])//这个商品是中间商品 
				{
					cnt--;
					vis[que.top()]--;
				}
				que.pop();//卖出 
				que.push(x);//还可以再买，
				//相当于最小价值的商品变成了x，
				//可以选择不卖，这样相当于没有买入x 
				vis[x]++;//该商品做了中间商
				//(这种价格做的中间商++) 
			}
		}
		printf("%lld %d\n",ans,cnt*2);
	}
}

HDU-6438-Buy and Resell（贪心，物品买卖）

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