HDU 6438 Buy and Resell

http://acm.hdu.edu.cn/showproblem.php?pid=6438

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

题意：给出n个数 a[i]，可以在某天买入x,并在某天卖出 y ，利润就是y -x ，问最大利润和操作次数

题解：维护一个最小价值的优先队列，每次买入，操作次数就加二（因为要卖出）。注意卖出的时候可能在当天卖出并不是最优，这时候就要再次push一个中间价，表示把之前累积的利润抵消掉。

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#include<iostream>
#include<algorithm>
#include<cstdio>
#include<stdio.h>
#include<string.h>
#include<queue>
#include<cmath>
#include<map>
#include<set>
#include<vector>
using namespace std;
#define inf 0x3f3f3f3f
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define mem(a,b) memset(a,b,sizeof(a));
#define lowbit(x)  x&-x;
typedef long long ll;
typedef unsigned long long ull;
const double eps = 1e-6;
const int maxn = 1e5+5;
const int mod = 1e9+7;
inline int read()
{int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;
}struct cmp
{bool operator()(const pair<int,int> p1, const pair<int,int> p2){if(p1.first == p2.first) return p1.second > p2.second;return p1.first > p2.first; }
};
priority_queue<pair<int,int>,vector<pair<int,int>>,cmp>Q;
int main() {int t;cin>>t;while(t--){while(!Q.empty()) Q.pop();int n;cin>>n;ll x;ll sum = 0,st = 0;for(int i = 1; i <= n; i++){cin>>x;if(Q.empty()){Q.push(make_pair(x,2)); //买入}else{auto it = Q.top();ll v = it.first;//cout<<v<<endl;if(x > v){Q.pop();sum += 1ll*(x-v);st += it.second;Q.push(make_pair(x,0)); //压入一个中间价}Q.push(make_pair(x,2));  //将当前价值买入}}printf("%lld %lld\n",sum,st);}
}

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