贪心POJ2393:Yogurt factory

Posted 2020-10-24 听风不成泣

Description

The cows have purchased a yogurt factory that makes world-famous Yucky Yogurt. Over the next N (1 <= N <= 10,000) weeks, the price of milk and labor will fluctuate weekly such that it will cost the company C_i (1 <= C_i <= 5,000) cents to produce one unit of yogurt in week i. Yucky‘s factory, being well-designed, can produce arbitrarily many units of yogurt each week. 

Yucky Yogurt owns a warehouse that can store unused yogurt at a constant fee of S (1 <= S <= 100) cents per unit of yogurt per week. Fortuitously, yogurt does not spoil. Yucky Yogurt‘s warehouse is enormous, so it can hold arbitrarily many units of yogurt. 

Yucky wants to find a way to make weekly deliveries of Y_i (0 <= Y_i <= 10,000) units of yogurt to its clientele (Y_i is the delivery quantity in week i). Help Yucky minimize its costs over the entire N-week period. Yogurt produced in week i, as well as any yogurt already in storage, can be used to meet Yucky‘s demand for that week.

Input

* Line 1: Two space-separated integers, N and S. 

* Lines 2..N+1: Line i+1 contains two space-separated integers: C_i and Y_i.

Output

* Line 1: Line 1 contains a single integer: the minimum total cost to satisfy the yogurt schedule. Note that the total might be too large for a 32-bit integer.

Sample Input

4 5
88 200
89 400
97 300
91 500

Sample Output

126900

Hint

OUTPUT DETAILS: 
In week 1, produce 200 units of yogurt and deliver all of it. In week 2, produce 700 units: deliver 400 units while storing 300 units. In week 3, deliver the 300 units that were stored. In week 4, produce and deliver 500 units.

 

题目大意：

牛们收购了一个奶酪工厂,接下来的N个星期里，牛奶价格和劳力价格不断起伏．第i周，生产一个单位奶酪需要Ci(1≤Ci≤5000)便士．工厂有一个货栈，保存一单位奶酪，每周需要S(1≤S≤100)便士，这个费用不会变化．货栈十分强大，可以存无限量的奶酪，而且保证它们不变质．工厂接到订单，在第i周需要交付Yi(0≤Yi≤104)单位的奶酪给委托人．第i周刚生产的奶酪，以及之前的存货，都可以作为产品交付．请帮牛们计算这段时间里完成任务的最小代价．

第1行输入两个整数N和S．接下来N行输入Ci和Yi．

输出最少的代价．注意，可能超过32位长整型

提示翻译：

第1周生产200单位奶酪并全部交付；第2周生产700单位，交付400单位，有300单位；第3周交
付300单位存货．第4周生产并交付500单位．

 

思路：更新下一周的生产成本即可

 

代码如下：

 1 #include <iostream>
 2 #include <cstdio>
 3 using namespace std;
 4 const int maxn = 10005;
 5 struct node
 6 {
 7     int c,y;
 8 };
 9 node a[maxn];
10 int main()
11 {
12     int n,s;
13     long long sum;
14     while(~scanf("%d%d",&n,&s))
15     {
16         for(int i=0;i<n;i++)
17         {
18             scanf("%d%d",&a[i].c,&a[i].y);
19         }
20         sum=0;
21         for(int i=0;i<n;i++)
22         {
23             sum+=a[i].c*a[i].y;
24             if(i!=n-1);
25             {
26                 a[i+1].c=min(a[i+1].c,a[i].c+s);
27             }
28         }
29         cout << sum << endl;
30     }
31 
32     return 0;
33 }

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