第九道ACM程序题

1.题目编号：1015

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
   
   
   
   
   
   
   
   

    
    
    
    2.简单题意：一个酸奶工厂，产生每一单位的奶需要花费c美分，酸奶不会腐烂，如果储存在仓库里。每单位将会花费固定的金额s美分，每周需要给客户提供y单位的酸奶，计算怎样才能使工厂在n个星期内花费最少
   
   
   
   
   
   
   
   

    
    
    
    3.解题思路形成过程：若要花费的少，则需要看看先给顾客以前生产的划算还是把刚刚生产出来的给顾客，因为以前生产的有固定s美分的花费，运用贪心算法。
   
   
   
   
   
   
   
   

    
    
    
    4.感悟：这个题目也是运用了贪心算法，最难想到的是给客户的酸奶是与前一周的酸奶有联系，搞清楚了之后，题目输出还说 Note that the total might be too  large for a 32-bit integer，可惜我的两个大眼睛没有看到，
    
    
    
    wa了很多次，runtime了很多次，但是最后能过还是万幸，多思考，多看题。。
   
   
   
   
   
   
   
   

    
    
    
    5.AC的代码：
   
   
   
   
   
   
   
   

    
    
    
    #include<iostream>
#include<stdio.h>
using namespace std;
   
   
   
   
   
   
   
   

    
    
    
    int main()
{
    long long n,s,totle=0;
    long long c[11000],y[11000];
   
   
   
   
   
   
   
   

    
    
    
       cin>>n>>s;
   
   
   
   
   
   
   
   

    
    
    
        for (int i=0;i<n;i++)
       cin>>c[i]>>y[i];
   
   
   
   
   
   
   
   

    
    
    
        for (int i=1;i<n;i++){
       if (c[i-1]+s<c[i])
        c[i]=c[i-1]+s;
        else
        c[i]=c[i];
    }
   
   
   
   
   
   
   
   

    
    
    
        for (int i=0;i<n;i++)
      totle+=c[i]*y[i];
       printf ("%I64d\n",totle);
   
   
   
   
   
   
   
   

    
    
    
        return 0;
}