Supermarket | Greedy

Supermarket | Greedy

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from poj 1456

from acwing 145

Time limit: 2s

Memory limit: 65M

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Description:

A supermarket has a set Prod of products on sale. It earns a profit px for each product x∈Prod sold by a deadline dx that is measured as an integral number of time units starting from the moment the sale begins. Each product takes precisely one unit of time for being sold. A selling schedule is an ordered subset of products Sell ≤ Prod such that the selling of each product x∈Sell, according to the ordering of Sell, completes before the deadline dx or just when dx expires. The profit of the selling schedule is Profit(Sell)=Σx∈Sellpx. An optimal selling schedule is a schedule with a maximum profit.
For example, consider the products Prod={a,b,c,d} with (pa,da)=(50,2), (pb,db)=(10,1), (pc,dc)=(20,2), and (pd,dd)=(30,1). The possible selling schedules are listed in table 1. For instance, the schedule Sell={d,a} shows that the selling of product d starts at time 0 and ends at time 1, while the selling of product a starts at time 1 and ends at time 2. Each of these products is sold by its deadline. Sell is the optimal schedule and its profit is 80.

Write a program that reads sets of products from an input text file and computes the profit of an optimal selling schedule for each set of products.

Input

A set of products starts with an integer 0 <= n <= 10000, which is the number of products in the set, and continues with n pairs pi di of integers, 1 <= pi <= 10000 and 1 <= di <= 10000, that designate the profit and the selling deadline of the i-th product. White spaces can occur freely in input. Input data terminate with an end of file and are guaranteed correct.

Output

For each set of products, the program prints on the standard output the profit of an optimal selling schedule for the set. Each result is printed from the beginning of a separate line.

Sample Input

4  50 2  10 1   20 2   30 1

7  20 1   2 1   10 3  100 2   8 2
   5 20  50 10

Sample Output

80
185

Hint

The sample input contains two product sets. The first set encodes the products from table 1. The second set is for 7 products. The profit of an optimal schedule for these products is 185.

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Main idea:

A supermarket has some products that are about to expire, and can only sell one product per day. Given the profit and expiration date of these expired products, ask you what is the maximum revenue that this supermarket can get.

When you look at this topic, you must think of greed the first time! ?

For pure greedy, if you don’t need priority queue or merge search optimization, then start with the highest value product, and each product is sold on the last day of expiration. If there are already higher-priced items that need to be sold that day, then Just traverse forward to find the day when the item can be sold.

AC code:

#include<iostream>
#include<algorithm>
using namespace std;
#define pii pair<int,int>
pii w[10005];               //存储商品信息
bool note[10005];           //用来标记当天是否已经有更高价格得商品需要出售
bool cmp(pii a,pii b){
    
    
    return a.first > b.first;   //贪心，按照利润从大到小排序
}
int n,max_day,sum;            //商品数，最大过期日期，最大利益
int main(){
    
    
    while(cin>>n){
    
    
        max_day = sum = 0;
        for(int i = 1;i <= n;++i)
            cin>>w[i].first>>w[i].second,max_day = max(max_day,w[i].second);
        sort(w + 1,w + 1 + n,cmp);
        for(int i = 1;i <= max_day;++i)
            note[i] = false;
        for(int i = 1;i <= n;++i){
    
    
            if(!note[w[i].second])      //如果该商品的最后一天没有更高利润的商品出售，那么就在当天卖出
                note[w[i].second] = true,sum += w[i].first;
            else{
    
                           //否则，往前查找可以用来卖该商品的日子
                int j = w[i].second;
                while(note[j])
                    --j;
                if(j)
                    note[j] = true,sum += w[i].first;
            }
        }
        cout<<sum<<"\n";
    }
    return 0;
}

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Other methods:
Greedy + Priority Queue
Greedy + Union Check