hdu6438（贪心）

题目：

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≤10^5)
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≤10^9)
It is guaranteed that the sum of all n is no more than 5×10^5.

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：

一个人一开始有无限多的钱，按时间顺序每天去一个地方，到了这个地方，他可以选择：

买这个地方的东西，卖手上的东西，什么都不做

但一个地方只能买进或卖出一件东西，问你一直到最后一个城市，最多能赚多少钱。

这个题第一眼看过去就是一道贪心，策略就是，假设把每天都买入，然后如果手上买入的最低价a小于当天的价格b就卖出，当天b依然买入，这样如果后面有更高的价格c，并且卖出了这天的商品b，那就相当于b作为中间状态过度掉了，这个时候交易次数不变，价格升高了。
后来看题解，发现这实际上是cf原题round#437 E，只不过加了个交易次数而已。

代码：

#include
using namespace std;

typedef long long ll;
priority_queuevector,greater > q;

int main()
{
        while(!q.empty())
            q.pop();
        int n;
        ll ans = 0;
        scanf("%d",&n);
        for(int i = 0;i < n;i++){
            ll a;
            scanf("%lld",&a);
            if(q.empty() || q.top() >= a)
                q.push(a);
            else{
                ans += a-q.top();
                ll temp = q.top();
                q.pop();
                q.push(a);
                q.push(a);          
            } 
        }
        printf("%lld\n",ans);
}