bzoj 16801740 : [Usaco2005 Mar]Yogurt factory 贪心 双倍经验
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1680: [Usaco2005 Mar]Yogurt factory
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.
吨酸奶。酸奶的生产受到很多因素的影响,所以每个月的生产成本是变化的,其中第 i 个月的成本是
每吨 Ci 元。
奶牛可以提前里把酸奶做好,存在仓库里,等需要的时候再拿出来卖。存储在仓库里的酸奶,每
吨酸奶存放一个月需要支付 S 元的维护费用,存放的时间可以任意长。假设工厂的产量是无限的,存
储酸奶的仓库也是无限大的。请问为了满足订单的需要,奶牛生产这些酸奶最少要花多少钱?
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.
• 单个整数:表示生产酸奶的最小总费用
Sample Input
88 200
89 400
97 300
91 500
Sample Output
OUTPUT DETAILS:
第一个月生产 200 吨酸奶;第二个月生产
700 吨酸奶,并存下 300 吨;第三个月不生产酸
奶;第三个月生产 500 吨
思路 :
每天选择最小价格做酸奶, 最小价格是之前的价格加上相应的储存价格和今天的价格取min
代码 :
#include <cstdio> #include <iostream> #include <cstring> #include <algorithm> using namespace std; int tmp; long long ans; struct node { long long a,c; }d[10001]; int main() { /*freopen("factory.in","r",stdin); freopen("factory.out","w",stdout);*/ int n,s; scanf("%d%d",&n,&s); scanf("%lld%lld",&d[1].c,&d[1].a); tmp=d[1].c; ans+=tmp*d[1].a; for(int i=2;i<=n;i++) { tmp+=s; scanf("%lld%lld",&d[i].c,&d[i].a); if(tmp>=d[i].c) tmp=d[i].c; ans+=tmp*d[i].a; } printf("%lld",ans); }
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