POJ 2393 Yogurt factory

Yogurt factory

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5472		Accepted: 2811

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. 

 

第i周生产奶的单价为C_i，每单位奶保存一周的花费为S，每周要供贷Y_i，求最终最小的成本。

采用贪心算法，每周将本周的生产成本与（上周所采取的生产成本+S）进行比较，取一个小的作为本周的生产成本

 

#include<iostream>
#include<cstdio>

using namespace std;

int n,s;

int main()
{
    int c,y,current;
    while(scanf("%d %d",&n,&s)==2)
    {
        long long ans=0;
        scanf("%d %d",&c,&y);
        current=c;
        ans+=current*y;
        for(int i=2;i<=n;i++)
        {
            scanf("%d %d",&c,&y);
            current+=s;
            if(current>c)
                current=c;
            ans+=current*y;
        }
        cout<<ans<<endl;
    }

    return 0;
}