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  • POJ 2393 Yogurt factory

    Yogurt factory
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 5472   Accepted: 2811

    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. 
     
    第i周生产奶的单价为C_i,每单位奶保存一周的花费为S,每周要供贷Y_i,求最终最小的成本。
    采用贪心算法,每周将本周的生产成本与(上周所采取的生产成本+S)进行比较,取一个小的作为本周的生产成本
     
     1 #include<iostream>
     2 #include<cstdio>
     3 
     4 using namespace std;
     5 
     6 int n,s;
     7 
     8 int main()
     9 {
    10     int c,y,current;
    11     while(scanf("%d %d",&n,&s)==2)
    12     {
    13         long long ans=0;
    14         scanf("%d %d",&c,&y);
    15         current=c;
    16         ans+=current*y;
    17         for(int i=2;i<=n;i++)
    18         {
    19             scanf("%d %d",&c,&y);
    20             current+=s;
    21             if(current>c)
    22                 current=c;
    23             ans+=current*y;
    24         }
    25         cout<<ans<<endl;
    26     }
    27 
    28     return 0;
    29 }
    [C++]
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  • 原文地址:https://www.cnblogs.com/lzj-0218/p/3246510.html
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