2018中国大学生程序设计竞赛

Buy and Resell

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1126    Accepted Submission(s): 359

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

 

题意：n个点，每个点都有自己的价值val[i]，我们可以在每个点买入和卖出，如在A点买入B点卖出，我的收益是val[B]-val[A]，怎样操作我的收益最大？

分析：在每一个点上进行卖出操作，将val[i]放进优先级队列q（已负数形式加入相当于最小的在最上面），则每一次我卖出的收益为val[i]-q.top()

　　　当前面没有比这个点价值小的点时我会卖出这个点，这样存入的点要买入都会先卖出一次然后才能在遇到更大的价值时再买入一次，所以每个点存入两次

　　　这样我可以求出所得的最大利益

　　　现在求最大利益下的买入卖出次数，对每个push进队列的数我们打一个标记，1代表第一次卖出这个数，0代表第二次买入这个数

　　　由上诉分析可以得到：当我买入一个数时，也就代表了我进行了买入卖出一共两次操作

AC代码：

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll maxn = 1e4+10;
const double eps = 1e-8;
const ll mod = 1e9 + 7;
const ll inf = 1e9;
const double pi = acos(-1.0);
priority_queue<pair<ll,ll> > q;
int main() {
    ll T, n;
    scanf("%lld",&T);
    while( T -- ) {
        scanf("%lld",&n);
        ll ans = 0, cnt = 0;
        while(!q.empty()) {
            q.pop();
        }
        for( ll i = 0, x; i < n; i ++ ) {
            scanf("%lld",&x);
            q.push(make_pair(-x,0)), q.push(make_pair(-x,1));
            ans += x + q.top().first;
            if( !q.top().second ) {
                cnt += 2;
            }
            q.pop();
        }
        printf("%lld %lld
",ans,cnt);
    }
    return 0;
}