Yogurt factory

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. 

 

 

题目意思：由于市场原因牛奶的生产价格会有所波动，要求在将来的n周里，该公司拥有一个足够大的仓库储存制好的牛奶，每个单位的牛奶每周储存

 所需的费用为s，现给出将来n周生产每单位牛奶的价格及客户所需的数量（即提前知道了下面几周的牛奶价格和客户需求！！！），求出该公司将来n周内所要支付的最小生产值。

 

解题思路：这是一道贪心题，贪心策略，每一个周只需要和下一个周对比，每一次更新下一个相邻的周，因为若下一个周被更新了，那么下一个周可以用来更新剩下的周，所以每一个周只需要对下一个周负责就可以了，可以看成一种带有传递性的比较。

 

上代码：

#include<stdio.h>
struct Yogurt
{
    int c;
    int y;
};
int main()
{
    int n,s,i;
    long long ans;
    struct Yogurt a[10010];
    a[0].c=999999;
    while(scanf("%d%d",&n,&s)!=EOF)
    {
        for(i=1; i<=n; i++)
        {
            scanf("%d%d",&a[i].c,&a[i].y);
        }
        for(i=1; i<=n; i++)
        {
            if(a[i].c<a[i-1].c+s)
                a[i].c=a[i].c;
            else
                a[i].c=a[i-1].c+s;
        }
        ans=0;
        for(i=1; i<=n; i++)
        {
            ans=ans+a[i].c*a[i].y;
        }
        printf("%lld\n",ans);
    }
    return 0;
}