• 【贪心】POJ2393:Yogurt factory


    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 }
    View Code
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  • 原文地址:https://www.cnblogs.com/SoulSecret/p/8454926.html
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