POJ2393 -- Yogurt factory -- 贪心

Yogurt factory

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11215 Accepted: 5697

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

4 588 20089 40097 30091 500

Sample Output

126900

Hint

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. 

Source

USACO 2005 March Gold

问题描述: 第一行给出n和s n代表星期,s代表每单位酸奶储存所需要的金额

下面给出n行 每行有两个整数 第一个整数代表该周生产1单位酸奶所需的金额 第二个整数代表该周酸奶的需求量

我们要求的是最优的生产策略的花费(如何使生产所有的酸奶花费最小)

分析:刚开始的时候准备维护一个优先队列 但是其实不需要使用单调队列 只要我们维护一个最小值即可

那么 每周我们都用我们可以使用金额的最小值来生产 就是最优策略 

而这个最小值为:

minPrice = min(minPrice + s , week[i]); (week[i]代表该周的金额)

比如本题案例

In week 2, produce 700 units: deliver 400 units while storing 300 units. In week 3, deliver the 300 units that were stored.

由于我们在第三周的价格大于第二周的价格 + 维护费用 所以我们应该在第二周就把第三周的酸奶生产好 然后支付这个维护费用即可

但我们可以换个角度 我们在第二周还是只生产第二周的酸奶 把这个维护费用加在第二周的生产价格上 作为第三周的生产价格 

就这样,我们在每周只生产自己的酸奶 只是使用最小的价格 最后就能得到最优的策略

代码:

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;priority_queue<int> q;int week[10000 + 5];int v[10000 + 5];int main(){int n,s;long long ans = 0;int amount = 0;int minPrice = 100000000;scanf("%d %d",&n,&s);for(int i = 0;i<n;i++){scanf("%d %d",&week[i],&v[i]);int pri = minPrice;minPrice = min(minPrice + s , week[i]);ans += (minPrice * v[i]);}printf("%lld\n",ans);}