SDNU——8题——D - Yogurt factory

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
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.

 #include<cstdio>
 #include<iostream>
 #include<vector>
 #include<cmath>
 using namespace std;
 int main()
 {
 	int n,s,i;
 	long long int ret=0;
 	cin>>n>>s;

 	int price=9999;
 	int pricenow,put;
 	for(i=0;i<n;i++)
 	{
 		cin>>pricenow>>put;
 		price=min(pricenow,price+s);
 		ret+=price*put;
	 }
	 cout<<ret;
 }

这个题让我学到了很多

题目大意：输入今天的产奶价格和最少产奶量（可以多产存下来，但是存的时候也要花钱），问怎么控制产奶量使得花费最少~~

我一开始~~莽夫暴力——把今天价格和以后所有的价格相比（有个if跳出），麻烦是麻烦，但是能实现也好啊，那个存下来的价格+产奶价格和下一次价格的比较我就没能完美实现，（可能有点乱，那就别看我这个暴力思路了）

接下来是贪心算法——这个思路太棒了——费用只用和下一天的比，而起比较的是单价即可（今日单价比昨日单价+存储单价）如果大……就把今日价格更新为昨日单价+存储单价，然后与数量相乘得出目前最少花费~~

虽然只考虑一个方面，但就是又数学依据的——1.我们可以一天一天的算出最优单价，和最优花费，而且，会进行价格更新，无非就是两种情况{1}今天要生产的在昨天生产更值一些，那就把今天生产的价格，更新为昨天的价格加上单价（因为已经存了一天了），如果不值，就不必更新，直接，算出今天花费即可。又因为价格是更新的，所以可能第一天就会生产3的内容，因为最优价格是跟着天数走的~~（好好想一想，我可能说的有点乱了，希望能表达出我的点）没必要捆在一起考虑

2.相邻两天考虑的是：昨日：（昨日单价+存贮单价）*明日数量与明日单价*明日数量==》一约分你就会发现需要比较的就是单价

这就很考验数学了QAQ~~