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  • POJ2393奶酪工厂

    Yogurt factory
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 14771   Accepted: 7437

    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. 

    Source

     
    ******每次判断哪个地方更适合做单个奶酪,就将从现在开始及以后都按这套方案走,如果遇到更优的方案当然就修改方案啦,这就是传说中的贪心啦。
     
     1 #include<cstdio>
     2 #include<cstring>
     3 #include<algorithm>
     4 #include<cmath>
     5 using namespace std;
     6 int i,j,n,c[10005],w[10005],s,k = 1;
     7 long long int ans = 0;
     8 int main()
     9 {
    10     scanf("%d %d",&n,&s);
    11     for(i = 1;i <= n;i++)
    12     {
    13         scanf("%d %d",&c[i],&w[i]);
    14         if(c[i] <= c[k] + (i - k) * s)
    15         k = i;
    16         ans += c[k] * w[i] + (i - k) * s * w[i];
    17     }
    18     printf("%lld",ans);
    19     return 0;
    20 }
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  • 原文地址:https://www.cnblogs.com/rax-/p/9857801.html
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