題目描述：

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 5
88 200
89 400
97 300
91 500

Sample Output

126900

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這道題是我在HURBST做的題目，連結是https://vjudge.net/contest/262212#problem/D

中文理解：有一個酸乳酪生產商，每一個星期都有其需求量和價格量，乳酪可以儲存到下個星期使用，不過有儲存費用，此題目要求你找出最小費用的解；

個人看法：這道題目是貪心的基礎題目，它滿足了貪心演算法的基於當前是最優解，找出下一個最優解，保證這一步是最優解。

 1 #include<stdio.h>
 2
 
 #include<algorithm>
 3 #include<math.h>
 4 #include<vector>
 5 #include<iterator>
 6 #include<string>
 7 #include<string.h>
 8 #include<ctype.h>
 9 #include<map>
10 #include<stack>
11 #include<queue>
12 #include<iostream>
13 #include<time.h>
14 
15 using namespace std;
16 
17 #define rep(i ,a, b) for(int i = a; i <= b; i++)
18 #define per(i, a, b) for(int i = a; i <= b; i--)
19 typedef long long ll;//資料量大，用longlong來進行運算較合適。不然會資料溢位。
20 
21 struct milk
22 {
23     ll c_i, y_i, id;
24 }a[10006];
25 
26 ll findmin(ll n, ll s)
27 {
28     ll ans=a[0].c_i*a[0].y_i, k = 0;
29     for(ll i = 1; i < n; i++)
30     {
31         ll temp = a[k].c_i+s*(i- a[k].id), cur = a[i].c_i;//記錄前一週牛奶的價格和這個一週的價格，進行比較然後，選取較小的作為當前的最小解；
32         if(temp < cur)
33         {
34             ans += temp*a[i].y_i;
35         }
36         else
37         {
38             ans += cur*a[i].y_i;
39             k = i;
40         }
41     }
42     return ans;
43 }
44 
45 int main()//這是貪心問題。
46 {
47     ll n, s;
48     while(scanf("%lld%lld", &n, &s) != EOF)
49     {
50         for(ll i =0; i < n; i++)
51         {
52             scanf("%lld%lld", &a[i].c_i, &a[i].y_i);
53             a[i].id = i;
54         }
55         printf("%lld\n", findmin(n, s));
56     }
57     return 0;
58 }

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