Codeforces 867 E. Buy Low Sell High （贪心+优先队列）

末蓝、 2021-12-05 15:11 235阅读 0赞

You can perfectly predict the price of a certain stock for the next N days. You would like to profit on this knowledge, but only want to transact one share of stock per day. That is, each day you will either buy one share, sell one share, or do nothing. Initially you own zero shares, and you cannot sell shares when you don’t own any. At the end of the N days you would like to again own zero shares, but want to have as much money as possible.

Input  
begins with an integer N (2 ≤ N ≤ 3·105), the number of days.  
Following this is a line with exactly N integers p1, p2, …, pN (1 ≤ pi ≤ 106). The price of one share of stock on the i-th day is given by pi.

Output  
Print the maximum amount of money you can end up with at the end of N days.

Examples  
Input  
9  
10 5 4 7 9 12 6 2 10  
Output  
20

Input  
20  
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4  
Output  
41

Note  
In the first example, buy a share at 5, buy another at 4, sell one at 9 and another at 12. Then buy at 2 and sell at 10. The total profit is  - 5 - 4 + 9 + 12 - 2 + 10 = 20.

题意：从左往右，买小的卖大的，问最大利益。

思路：利用优先队列，每输入一个数就和队首比较并且存入队列，如果队首小，就说明可以获得利益，用一个sum存起来。注意，此时需要把输入的数在此放入队列，因为后面可能还有一个数可以跟队首获取更大的利益，1 2 和1 7 显然1 7 可以获得的利益大，那么把2在次放入队列让2 7 获得5的利益 + 1 2 获得1的利益=1 7直接获得的利益。

#include<iostream>
    #include<queue>
    using namespace std;
    typedef long long ll;
    priority_queue<ll,vector<ll>,greater<ll> > q;
    int main()
    {
    	ll n,ans;
    	while(cin>>n)
    	{
    		ans=0;
    		while(!q.empty())	q.pop();
    		while(n--)
    		{
    			ll a;
    			cin>>a;
    			q.push(a);
    			if(a>q.top())
    			{
    				ans+=a-q.top();
    				q.pop();
    				q.push(a);
    			}
    		}
    		cout<<ans<<endl;
    	}
    	return 0;
    }