HDU6348 Buy and Resell(2018CCPC网络赛，优先队列，贪心)

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

思路

本题和cf的一道题基本一样CodeForces 867E Buy Low Sell High(优先队列，贪心)。有n个城市，你需要进行一种货物交易，每个城市的价格不一样，在每个城市只能进行一次交易，问你如何交易可以使得最后的利润最大，求出最大值和最小的交易次数。

最大值的求法就是和cf那道题一样，用一个从小到大的优先队列维护，在每次进行交易的时候交易的次数就++，要使交易次数变少，那么就使得中间的变化次数最大，标记一下每个物品买入的价格，出现过就减少次数，没有就增加.

代码

#include 
using namespace std;
#define mem(a, b) memset(a, b, sizeof(a))
typedef long long ll;
const ll N = 1e5 + 40;
priority_queuevector, greater> q;
unordered_map mp;
void solve()
{
    while (!q.empty())
        q.pop();
    mp.clear();
    ll n, x, ans = 0, cnt = 0;
    scanf("%lld", &n);
    for (ll i = 1; i <= n; i++)
    {
        scanf("%lld", &x);
        if (q.empty() || x <= q.top())
        {
            q.push(x);
        }
        else
        {
            ans += x - q.top();
            cnt++;
            if (mp[q.top()])
            {
                cnt--;
                mp[q.top()]--;
            }
            q.pop();
            q.push(x), q.push(x);
            mp[x]++;
        }
    }
    printf("%lld %lld\n", ans, cnt * 2);
}
int main()
{
    // freopen("in.txt", "r", stdin);
    ll t;
    scanf("%lld", &t);
    while (t--)
        solve();

    return 0;
}