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

ゝ一纸荒年。 2021-11-17 16:22 464阅读 0赞

## 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; 
    }

转载于:https://www.cnblogs.com/guangheli/p/10972547.html