[数据结构与算法]Buy and Resell

Buy and Resell

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.

题意：有 n 天，给出每天物品的价格，一个人最初有无限多的钱，每天可以选择买一个物品、卖一个物品或者什么都不做，问这个人的最大收益。

题解：用优先队列维护一下价格查一下最大的收益差

#include<iostream>
#include<stack>
#include<queue>
#include<map>
#include<algorithm>
#include<string>


using namespace std;

const int N = 1e5 + 10;

int main(void) {
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        long long ans = 0, x;
        map<int, int> m;
        long long cnt = 0;
        priority_queue<int, vector<int>, greater<int> > q;
        for (int i = 0; i < n; i++) {
            cin >> x;
            q.push(x);
            if (q.top() < x) {
                ans += x - q.top();
                cnt++;

                if (m[q.top()]) {
                    m[q.top()]--;
                    cnt--;
                }
                m[x]++, q.pop();
                q.push(x);
            }
        }
        cout << ans << " " << cnt * 2 << endl;
    }
    return 0;
}