﻿ Buy and Resell | 玄奇博客 ### 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

### Source

2018中国大学生程序设计竞赛 – 网络选拔赛

chendu

### 解法

1. 替代前面已经匹配的一对的后一个，次数不需要增加，只增加值
2. 与前面没有匹配且比当前值小的最小的进行匹配，次数增加，值增加二者差值。

### 代码

``````#include
#include
#include
#include

using namespace std;

int main(){
//freopen("in.txt", "r", stdin);
int T;
scanf("%d", &T);
while(T--){
int n;
scanf("%d", &n);
priority_queue , greater > quedengdai, yipipei;

while(!quedengdai.empty())quedengdai.pop();  //清空队列
while(!yipipei.empty())yipipei.pop();

long long ans=0;
int tot=0;
for(int i=1; i<=n; i++){
int num;
scanf("%d", &num);

int num1=-1,num2=-1;

if(!yipipei.empty()) num1 = num - yipipei.top();  //计算替代和新匹配的哪个增加的值多选哪个

if(!quedengdai.empty()) num2 = num - quedengdai.top();

if(!yipipei.empty() && yipipei.top() < num && num1 >= num2){  //替换已匹配

ans += num - yipipei.top();

quedengdai.push(yipipei.top());

yipipei.pop();

yipipei.push(num);
}
else {  //可能为空
if(!quedengdai.empty() && quedengdai.top() < num){  //与未匹配的匹配

ans += num - quedengdai.top();

tot+=2;

quedengdai.pop();

yipipei.push(num);

//                  printf("two:");
}
else {  //放入等待匹配队列

quedengdai.push(num);
}
}
}
printf("%lld %d\n", ans,tot);
}
}

``````

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