nyoj 208 + poj 1456 Supermarket (贪心)

Supermarket

时间限制：1000 ms  |  内存限制：65535 KB

难度：4

 

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.

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

 /**
     题目大意：
         求：能够得到的最大利润
         数据说明：
             n ==> 表示有n种商品
             px dx ==> 只要商品在dx （包括dx）之前买完就可以获得px的利润
             一天只可以买一种商品

     步骤（贪心）：
         ①、将px降序排列
         ②、每件商品从最大期限的时间的开始查找直到找到可以使用的时间 （或者到第一天）
             ②（1）如果是找到满足条件的时间就将该件商品的px加入能够得到的总利润中（然后跳出循环）
             ②（2）如果是找不到就不将该商品的px加入到总利润中
 **/

核心代码：

 sort (P, P+n, cmp);

 for (int i = ; i < n; ++ i)
 {
     for (int j = P[i].dx; j >= ; -- j)
     {
         if (!book [i])
         {
             book [i] = ;
             cnt += P[i].px;
             break;
         }
     }
 }

C/C++代码实现（AC）：

 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 #include<iostream>

 using namespace std;

 int n, cnt, flag[*];

 struct node
 {
     int v, k;
 }P[*];

 bool cmp(node a, node b)
 {
     return a.v > b.v;
 }

 int main(){
     while(~scanf("%d", &n))
     {
         cnt = ;
         memset(flag, , sizeof(flag));
         for(int i = ; i < n; ++ i)
             scanf("%d%d", &P[i].v, &P[i].k);
         sort(P, P + n, cmp);
         for(int i = ; i < n; ++ i)
         {
             for(int j = P[i].k; j >= ; -- j)
             {
                 if (!flag[j])
                 {
                     flag[j] = ;
                     cnt += P[i].v;
                     break;
                 }
             }
         }
         printf("%d\n", cnt);
     }
     return ;
 }