acm 1015 最佳生产方式

1.1015

2.

Problem 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. <br> <br>* 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

3.由生产成本和储存成本计算最佳生产方式

4.最优单价：本周最优是上周最优价加上储存成本和这周生产单价取 min      g[i] = min ( g[i-1]+s , c[i] ) ，这是我从xue博客上看的TVT

5.

#include<iostream>

#include<algorithm>

#include<map>

#include<set>

#include<numeric>

#include<math.h>

#include<string.h>

using namespacestd;

int a[10009];

int b[10009];

int main()

{  

    long long sum=0;

    int n,s,i;

    cin>>n>>s;

    for(i=1;i<=n;i++)

    {

        cin>>a[i];

        cin>>b[i];

        if(i==1)

        {

            sum+=a[i]*b[i];

            continue;

        }

        if(a[i-1]+s<a[i])sum+=(a[i-1]+s)*b[i];

        else sum+=a[i]*b[i];

       

    }

    cout<<sum<<endl;

    return 0;

}