1680: [Usaco2005 Mar]Yogurt factory
Time Limit: 5 Sec Memory Limit: 64 MB
Submit: 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
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）
#include<cstdio>
inline int read()
{
int x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
int n,s,save=;
long long ans;
int main()
{
n=read();
s=read();
for (int i=;i<=n;i++)
{
int w=read(),c=read();
save+=s;
if (save>w)save=w;
ans+=(long long)save*c;
}
printf("%lld",ans);
}