# BZOJ1680: [Usaco2005 Mar]Yogurt factory

## 1680: [Usaco2005 Mar]Yogurt factory

Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 106 Solved: 74

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## 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

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.

## HINT

## Source

题解：

刚看完题目，卧槽，这不是费用流吗？n=10000，瞬间蔫了。。。。。。。。

刚看完题目，卧槽，这不是费用流吗？n=10000，瞬间蔫了。。。。。。。。

原来是贪心：

对于每一天可以从当天转移也可以从以前转移，只要保存一个以前的最小值，在这一天加一个s再跟c比较，更新完最小值直接更新答案。

水爆了。。。

代码：（copy）

1 #include<cstdio> 2 inline int read() 3 { 4 int x=0,f=1;char ch=getchar(); 5 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 6 while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 7 return x*f; 8 } 9 int n,s,save=100000000; 10 long long ans; 11 int main() 12 { 13 n=read(); 14 s=read(); 15 for (int i=1;i<=n;i++) 16 { 17 int w=read(),c=read(); 18 save+=s; 19 if (save>w)save=w; 20 ans+=(long long)save*c; 21 } 22 printf("%lld",ans); 23 }