贪心的一道题 Supermarket

痛定思痛。 2022-06-13 09:19 154阅读 0赞

POJ 1456

描述

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)=Σ x∈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.  
![1456_1.jpg][]  
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.

输入

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.

输出

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.

样例输入

4  50 2  10 1   20 2   30 1
    
    7  20 1   2 1   10 3  100 2   8 2
       5 20  50 10

样例输出

80
    185

提示

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.

来源

Southeastern Europe 2003

题意大概是这样的

对于每个商品， 给出它的价值和它的截止时间， 只有在截止时间结束之前把商品卖掉才可以收获相应的价值

那么如何合理的安排天数才能够使得利益最大化

这个明显是一道贪心的问题，我的做法很简单，将物品按照价值从小到大排序，然后对于价值高的物品优先安排，看其能不能放到它的ddl那天，如果不能放到ddl那天，说明被占用的天被用来卖价值更高的物品了，那么我们就从ddl倒着向前找，找到没有被占用的一天把它占用掉

代码如下，PS今天懒得看并查集，网上的题解说可以用并查集优化，即不是一个个向前找，直接找一下pre就可以在o(1)的时间内找到可以用的天数

反正懒得写了

// ConsoleApplication1.cpp : 定义控制台应用程序的入口点。  
//  
  
  
\#include "stdafx.h"  
\#include<stdio.h>  
\#include<iostream>  
\#include<algorithm>  
\#include<memory.h>  
using namespace std;  
int n;  
struct node  
\{  
int val, ddl;  
bool operator < (const node & n1)  
\{  
return val > n1.val;  
\}  
\}p\[10010\];  
bool d\[10010\];  
int main()  
\{  
while (cin >> n)  
\{  
long res = 0;  
for (int i = 0; i < n; i++)  
scanf("%d %d", &p\[i\].val, &p\[i\].ddl);  
sort(p, p + n);  
memset(d, true, sizeof(d));  
for (int i = 0; i < n; i++)  
\{  
if (d\[p\[i\].ddl\])  
\{  
d\[p\[i\].ddl\] = false;  
res += p\[i\].val;  
\}  
else  
\{  
for (int j = p\[i\].ddl - 1; j > 0; j--)  
\{  
if (d\[j\])  
\{  
d\[j\] = false;  
res += p\[i\].val;  
break;  
\}  
\}  
\}  
\}  
printf("%d\\n",res);//cout << res << endl;  
\}  
  
  
  
return 0;  
\}

[1456_1.jpg]: /images/20220613/3d75f3f586004d30bae6c564e5e04ab3.png