BZOJ 1740: [Usaco2005 mar]Yogurt factory 奶酪工厂 贪心 + 问题转化

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.

 牛们收购了一个奶酪工厂,接下来的N个星期里，牛奶价格和劳力价格不断起伏．第i周，生产一个单位奶酪需要Ci(1≤Ci≤5000)便士．工厂有一个货栈，保存一单位奶酪，每周需要S(1≤S≤100)便士，这个费用不会变化．货栈十分强大，可以存无限量的奶酪，而且保证它们不变质．工厂接到订单，在第i周需要交付Yi(0≤Yi≤104)单位的奶酪给委托人．第i周刚生产的奶酪，以及之前的存货，都可以作为产品交付．请帮牛们计算这段时间里完成任务的最小代价．

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.

 

    第1行输入两个整数N和S．接下来N行输入Ci和Yi．

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.

    输出最少的代价．注意，可能超过32位长整型．

题解：

太经典了.

不难贪心证明，每次只可能用同一价格的存储方式.

我们只需维护一个 $minv$ ，表示价格最小值.

到下一周时，用 $minv[cur-1]$ 与 $cur_cost$ 作比较，并更新 $minv[i]$ 即可.

甚至都不用数组，几个变量即可搞定.

#include<bits/stdc++.h>
#define setIO(s) freopen(s".in","r",stdin)
#define ll long long
using namespace std;
int main()
{
	// setIO("input");
	int n;
	ll s,c,y,cur=0,ans=0;
	scanf("%d%lld",&n,&s);
	scanf("%lld%lld",&c,&y);
	ans+=c*y, cur=c;
	for(int i=2;i<=n;++i)
	{
		scanf("%lld%lld",&c,&y);
		cur=min(cur+s,c);
		ans+=cur*y;
	}
	printf("%lld\n",ans);
	return 0;
}