Supermarket

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∈Sell

px. 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.

輸入

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.

輸出

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.

樣例輸入

4  50 2  10 1   20 2   30 1

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

樣例輸出

80
185

 1 /**
 2     題目大意：
 3         求：能夠得到的最大利潤
 4         數據說明：
 5             n ==> 表示有n種商品
 6             px dx ==> 只要商品在dx （包括dx）之前買完就可以獲得px的利潤
 7             一天只可以買一種商品
 8             
 9     步驟（貪心）：
10         ①、將px降序排列
11         ②、每件商品從最大期限的時間的開始查找直到找到可以使用的時間 （或者到第一天）
12             ②（1）如果是找到滿足條件的時間就將該件商品的px加入能夠得到的總利潤中（然後跳出循環）
13             ②（2）如果是找不到就不將該商品的px加入到總利潤中
14 **/

核心代碼：

 1 sort (P, P+n, cmp);
 2 
 3 for (int i = 0; i < n; ++ i)
 4 {
 5     for (int j = P[i].dx; j >= 1; -- j)
 6     {
 7         if (!book [i])
 8         {
 9             book [i] = 1;
10             cnt += P[i].px;
11             break;
12         }
13     }
14 }

C/C++代碼實現（AC）：

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<cmath>
 4 #include<algorithm>
 5 #include<iostream>
 6 
 7 using namespace std;
 8 
 9 int n, cnt, flag[100*105];
10 
11 struct node
12 {
13     int v, k;
14 }P[100*105];
15 
16 bool cmp(node a, node b)
17 {
18     return a.v > b.v;
19 }
20 
21 int main(){
22     while(~scanf("%d", &n))
23     {
24         cnt = 0;
25         memset(flag, 0, sizeof(flag));
26         for(int i = 0; i < n; ++ i)
27             scanf("%d%d", &P[i].v, &P[i].k);
28         sort(P, P + n, cmp);
29         for(int i = 0; i < n; ++ i)
30         {
31             for(int j = P[i].k; j >= 1; -- j)
32             {
33                 if (!flag[j])
34                 {
35                     flag[j] = 1;
36                     cnt += P[i].v;
37                     break;
38                 }
39             }
40         }
41         printf("%d\n", cnt);
42     }
43     return 0;
44 }

nyoj 208 + poj 1456 Supermarket (貪心)