HDU 6438 Buy and Resell 【贪心+思维+优先队列】

传送门：HDU 6438

Buy and Resell Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

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
3
4
1 2 10 9
5
9 5 9 10 5
2
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

题意：

有n个城市，在第i个城市中，你有三种策略，你可以用ai元购买一个物品，或者用ai元卖掉一个物品，或者不买也不卖，你可以同时保存多个物品。最开始你身上没有物品，但是又无限的金钱，现在要求你从城市1走到城市n，问你最大的收益是多少,同时交易次数是多少。

题解：

很容易想到的是，我们应该花较少的钱买入，然后卖出更多的钱。如果我们在第i天买入，第j天卖出那么必须满足a[j]>a[i]a[j]>a[i]a[j]>a[i]，收入为a[j]−a[i]a[j]-a[i]a[j]−a[i],如果后面存在第k天，使得a[k]>a[j]a[k]>a[j]a[k]>a[j]，那么我们就应该在第k天卖出物品，因为收益更大，为a[k]−a[i]a[k]-a[i]a[k]−a[i]。
我们发现有恒等式：a[k]−a[i]=a[k]−a[j]+a[j]−a[i]a[k]-a[i]=a[k]-a[j]+a[j]-a[i]a[k]−a[i]=a[k]−a[j]+a[j]−a[i]; 这说明第j天有没有卖出并不会影响最终结果（相当于中间商，且无差价）
但是我们又考虑到，第j天虽然有没有卖出不影响结果，但是有没有买入是影响结果的，所有我们还要考虑买入的情况。

用优先队列（小根堆）来实现上述过程，如果 j 的中间商，那么我们就往队列里存两次，一次是中间商（卖出），一次是买入。当当前的a大于队首元素时，我们就进行交易，同时如果队首元素是中间商的话交易次数是不改变的。

AC代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#include<map>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
const int N=110000;
int t,n;
ll cnt,ans;
priority_queue<ll,vector<ll>,greater<ll> >q;
map<ll,int> mp;
int main()
{ll a;scanf("%d",&t);while(t--){while(!q.empty()) q.pop();mp.clear();cnt=0,ans=0;scanf("%d",&n);for(int i=0;i<n;i++){scanf("%I64d",&a);if(!q.empty()&&q.top()<a){ll tmp=q.top();q.pop();ans+=a-tmp;q.push(a);//两次存入队列cnt++;mp[a]++;if(mp[tmp]!=0)//中间商{cnt--;//交易次数不改变mp[tmp]--;}}q.push(a);}printf("%I64d %I64d\n",ans,cnt*2);}return 0;
}

HDU 6438 Buy and Resell 【贪心+思维+优先队列】相关推荐

HDU - 6438 Buy and Resell（思维+ 贪心）
题目链接题意 t组输入,每组n个城市,给出每个城市的货物的价格,在每个城市可以进行三个操作:购买货物,卖出货物,不做操作,问从城市1到城市n(不能返回)最多能获得多少利润,且操作数最少(买和卖分别算 ...
HDU 6438 Buy and Resell【贪心】
http://acm.hdu.edu.cn/showproblem.php?pid=6438 Problem Description The Power Cube is used as a stash ...
Buy and Resell(贪心好题！)
这个贪心可以说是很巧妙了. Buy and Resell Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K ( ...
Buy and Resell HDU - 6438（补）更新贪心
The Power Cube is used as a stash of Exotic Power. There are nn cities numbered 1,2,-,n1,2,-,n where ...
HDU - 6438（贪心+思维）
链接:HDU - 6438 题意:给出 n ,表示 n 天.给出 n 个数,a[i] 表示第 i 天,物品的价格是多少.每天可以选择买一个物品,或者卖一个已有物品,也可以什么都不做,问最后最大能赚多少 ...
hdu6438 Buy and Resell 买卖物品 ccpc网络赛贪心
题目传送门题目描述: 有n座城市,每座城市都可以对一个物品进行一次的买进或者卖出,可以同时拥有多个物品,计算利润最大值,并且交易次数要最少.(买入卖出算两次操作) 思路: 建立两个小根堆优先队列, ...
HDU6348 Buy and Resell
Bryce1010模板 HDU6348 Buy and Resell 题意: 从前往后开始旅游,在每个村庄可以选择三个操作: (1)买入物品 (2)卖出物品 (3)不买不卖求最后的最大获益. 思路: ...
Buy and Resell
题目链接: Buy and Resell 大致题意: 有n个城市, 在每个城市中你可以选择花费a[i]的价格买一个能量块, 也可以选择以a[i]的价格卖出一个能量块(前提是你要有能量块可以卖). 你会 ...
【牛客每日一题】tokitsukaze and Soldier 题目精讲贪心、优先队列、堆
链接:https://ac.nowcoder.com/acm/problem/50439 来源:牛客网 ACM在线模板今天才发现牛客推出了一个每日一题的版块,3月25号就开始了,今天才发现,赶紧补救 ...
HDU 2795 Billboard (线段树+贪心)
HDU 2795 Billboard (线段树+贪心) 手动博客搬家:本文发表于20170822 21:30:17, 原地址https://blog.csdn.net/suncongbo/articl ...

HDU 6438 Buy and Resell 【贪心+思维+优先队列】

HDU 6438 Buy and Resell 【贪心+思维+优先队列】相关推荐

最新文章

热门文章