HDU 6438 Buy and Resell 【贪心】【优先队列】【map】

最新推荐文章于 2020-10-31 21:33:16 发布

原创最新推荐文章于 2020-10-31 21:33:16 发布 · 194 阅读

0 ·

CC 4.0 BY-SA版权

STL 同时被 2 个专栏收录

8 篇文章

订阅专栏

贪心

7 篇文章

订阅专栏

博客围绕Power Cube的买卖问题展开，给定多个城市的交易价格，商人可选择买入、卖出或不操作。需计算最大利润及获取最大利润的最少交易次数，还给出了输入输出格式和示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Buy and Resell

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1353 Accepted Submission(s): 447

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

#include<bits/stdc++.h>
using namespace std;
#define ll long long
map <int, int> vis;
int main(){
    int N;
    priority_queue <int, vector <int>, greater <int> > q;
    for(scanf("%d", &N); N; N--){
        int n;
        scanf("%d", &n);
        vis.clear();
        ll ans = 0;
        int cnt = 0;
        while(!q.empty()) q.pop();
        for(int i = 0, x; i < n; i++){
            scanf("%d", &x);
            if(!q.empty() && q.top() < x){
                ans += (x - q.top());
                cnt++;
                if(vis[q.top()] > 0) cnt--, vis[q.top()]--;
                q.pop();
                q.push(x);
                vis[x]++;
            }
            q.push(x);
        }
        printf("%lld %d\n", ans, cnt * 2);
    }
    return 0;
}