poj 2393 Yogurt Factory（贪心）

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. 

题意：工厂每周生产奶酪的成本不同，每周生产奶酪达到指标后多余的可以储存，储存需要每周S的费用，问怎么生产成本最低，输出最低成本。

思路：DP；

具体：用每周的成本和前几周的最低的储存成本比较。

ac代码：

#include<stdio.h>
#include<string.h>
int min(int a,int b)
{
    if(a>b)
    return b;
    else
    return a;
}
int main()
{
    int a,b,i,j,n,x[100000],y[100000];
    while(scanf("%d%d",&a,&b)!=EOF)
    {
        long long sum=0;
        for(i=1;i<=a;i++)
        scanf("%d%d",&x[i],&y[i]);
        for(i=2;i<=a;i++)
        x[i]=min(x[i],x[i-1]+b);
        for(i=1;i<=a;i++)
        sum+=x[i]*y[i];
        printf("%lld
",sum);
    }
}

有什么问题可以评论提问。