贪心算法HURUST题目

题目描述：

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

-----------------------
这道题是我在HURBST做的题目，链接是https://vjudge.net/contest/262212#problem/D

中文理解：有一个酸奶酪生产商，每一个星期都有其需求量和价格量，奶酪可以保存到下个星期使用，不过有保存费用，此题目要求你找出最小费用的解；

个人看法：这道题目是贪心的基础题目，它满足了贪心算法的基于当前是最优解，找出下一个最优解，保证这一步是最优解。

 #include<stdio.h>
 #include<algorithm>
 #include<math.h>
 #include<vector>
 #include<iterator>
 #include<string>
 #include<string.h>
 #include<ctype.h>
 #include<map>
 #include<stack>
 #include<queue>
 #include<iostream>
 #include<time.h>

 using namespace std;

 #define rep(i ,a, b) for(int i = a; i <= b; i++)
 #define per(i, a, b) for(int i = a; i <= b; i--)
 typedef long long ll;//数据量大，用longlong来进行运算较合适。不然会数据溢出。

 struct milk
 {
     ll c_i, y_i, id;
 }a[];

 ll findmin(ll n, ll s)
 {
     ll ans=a[].c_i*a[].y_i, k = ;
     for(ll i = ; i < n; i++)
     {
         ll temp = a[k].c_i+s*(i- a[k].id), cur = a[i].c_i;//记录前一周牛奶的价格和这个一周的价格，进行比较然后，选取较小的作为当前的最小解；
         if(temp < cur)
         {
             ans += temp*a[i].y_i;
         }
         else
         {
             ans += cur*a[i].y_i;
             k = i;
         }
     }
     return ans;
 }

 int main()//这是贪心问题。
 {
     ll n, s;
     while(scanf("%lld%lld", &n, &s) != EOF)
     {
         for(ll i =; i < n; i++)
         {
             scanf("%lld%lld", &a[i].c_i, &a[i].y_i);
             a[i].id = i;
         }
         printf("%lld\n", findmin(n, s));
     }
     return ;
 }