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POJ 2393 Yogurt factory

姬康平
2023-12-01
Yogurt factory
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 7418 Accepted: 3801

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.


只计算邻周的最小值肯定是不对的,原来想的是,每一周的生产量和之前所有周的生产量的花费相比,选择单价最小的,然后再一乘,得到最终最小的花费。

但是这样超时了,两个for循环,o(n方)的时间复杂度

然后开始优化,

如果隔得周越多,那么就会产生s的倍数的存储费用,如果隔得周数超过了每个周的存储费用,就没必要再往前找合适的周来生产牛奶了,因为这时的存储费用已经超过了之前要节省的费用。再往前找就没有意义了。

代码如下:



#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
struct week
{
    __int64 danjia;
    __int64 number;
}week[10005];
int main(void)
{
   // freopen("B.txt","r",stdin);
    int n;
    __int64 s;
    scanf("%d%I64d",&n,&s);
    for(int i=1;i<=n;i++)
    {
        scanf("%I64d%I64d",&week[i].danjia,&week[i].number);
    }
    __int64 sum=week[1].danjia*week[1].number;
    for(int i=2;i<=n;i++)
    {
        __int64 minx=week[i].danjia;
        for(int j=i-1;i-j<s&&j>=1;j--)
        {
            if(week[i].danjia>(week[j].danjia+s*(i-j)))
                minx=min(minx,(week[j].danjia+s*(i-j)));
        }
        sum+=minx*week[i].number;
    }
    printf("%I64d\n",sum);

    return 0;
}

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