本文共 2181 字，大约阅读时间需要 7 分钟。

Yogurt factory

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

4 588 20089 40097 30091 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.

题意：给出每周牛奶生产的单价和需求量，牛奶可以生产过量并保存下来，但是需要额外交保存费，求最小需要多少钱。

解题思路：若使用之前留下来的牛奶，那么第i周和第j周牛奶单价的差价就是(j-i)*s，可以从前往后递推，保存当前周牛奶的最小价，即a[i]=min(a[i],a[i-1]+s);根据递推式可以省去和更前面的比较，只需要与上一周比较。

#include 
   
    #include 
    
     #include 
     
      #include 
      
       using namespace std;const int maxn = 1e4 + 5;int n, s, c[maxn], y[maxn];int main() {    scanf("%d%d", &n, &s);    for (int i = 0; i < n; i++) scanf("%d%d", c + i, y + i);    for (int i = 1; i < n; i++) c[i] = min(c[i], c[i - 1] + s);    long long ans = 0;    for (int i = 0; i < n; i++) ans += c[i] * y[i];    printf("%lld\n", ans);    return 0;}
      
     
    
   

 

转载地址：https://blog.csdn.net/qq_44632981/article/details/98999113 如侵犯您的版权，请留言回复原文章的地址，我们会给您删除此文章，给您带来不便请您谅解！