Supermarket | 贪心 + 并查集

from poj 1456

from acwing 145

时间限制：2s

内存限制：65M

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 17  20 1   2 1   10 3  100 2   8 25 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.

题目大意：

一个超市有一些即将过期商品，且每天只能卖出一件商品，给定你这些过期商品的利润和过期日期，问你这家超市能获得的最多收益是多少。

这个题目一看肯定第一时间就是想到贪心吧！？

相比于纯贪心，利用并查集能做出一些优化，就是把每一天都指向这天之前（包括这天）的某一天能卖商品，比如第4、5、6天都有商品了，如果这时候有个商品也是在第六天过期，那么纯贪心，就需要遍历5，4直到3，而并查集就是直接将6指向3。

AC代码：

#include<iostream>
#include<algorithm>
using namespace std;
#define pii pair<int,int>
pii w[10005];               //存储商品信息
int father[10005];          //并查集用来存该商品最晚能卖的那天
int find(int x){if(father[x] == x)      //直到找到了当天还没有商品需要卖出的那天return x;return father[x] = find(father[x]);
}
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)father[i] = i;for(int i = 1;i <= n;++i){int d = find(w[i].second);if(d){father[d] = d - 1;            //该天卖了过后就不能再卖了，于是指向前一天sum += w[i].first;}}cout<<sum<<"\n";}return 0;
}

其他方法：
纯贪心
贪心 + 优先队列