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  • POJ 2393 Yogurt factory

    Yogurt factory
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 9466   Accepted: 4819

    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

     
     
        对于第 i 周(i>1),满足酸奶供应,这条最低的单位成本(包括生成成本和存储成本)为:min(本周的单位成本Ci , 前一周的最低单位成本+存储费用S)。
     1 #include <cstdio>
     2 #include <cstring>
     3 using namespace std;
     4 const int maxn=10005;
     5 int c[maxn],rc[maxn];
     6 long long y[maxn],sum;
     7 int min(const int &x,const int &y){return x<y?x:y;}
     8 int main()
     9 {
    10     int n,s,cost;
    11     while(scanf("%d%d",&n,&s)==2)
    12     {
    13         sum=0;
    14         for(int i=0;i<n;i++)
    15             scanf("%d%d",&c[i],&y[i]);
    16         memcpy(rc,c,sizeof(rc));
    17         rc[0]=c[0];
    18         for(int i=0;i<n;i++){
    19             cost=rc[i]=min(rc[i],c[i]);
    20             for(int j=i+1;j<n;j++){
    21                 cost+=s;
    22                 if(cost<c[j]) rc[j]=cost;
    23                 else break;
    24             }
    25         }
    26         for(int i=0;i<n;i++)
    27             sum+=y[i]*rc[i];
    28         printf("%lld
    ",sum);
    29     }
    30 }
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  • 原文地址:https://www.cnblogs.com/cumulonimbus/p/5890904.html
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