POJ 2393 Yogurt factory(贪心)

[align=center]Yogurt factory[/align]

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8091		Accepted: 4139

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：一家奶酪加工厂在n个星期里每个星期要为客户生产y_i单位的奶酪，每个星期里每单位的奶酪成本为c_i。工厂有一个巨大的（内存不限）仓库，每单位的储存费用为s。问完成n个星期的生产，最低费用是多少。

题解：找到每个星期要生产的奶酪的最低成本即可。若当前的 c_j 大于 (j-i)*s+c_i 则可以在之前就生产好，储存在仓库中。

蒟蒻的我写了一个O(n^2)算法，TLE(；′⌒`)  只会暴力，蠢到无可救药。

O(n)做法很简单，一直记录当前要生产的奶酪的最低生产价格就行了。

代码如下：

#include<cstdio>
#define ll long long
int main()
{
int n,s,c,y,c_min=9999;
while(scanf("%d%d",&n,&s)!=EOF)
{
ll ans=0;
while(n--)
{
scanf("%d%d",&c,&y);
if(c>c_min+s)
c=c_min+s;
c_min=c;
ans+=c*y;
}
printf("%lld\n",ans);
}
return 0;
}