POJ 2393 Yogurt factory 贪心

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 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.

题目大意:
有一个酸奶工厂向客户供奶，一共有n周，每周生产酸奶的单位价格都会变化，该工厂有一个仓库可以保存酸奶，每保存一周有固定的保存费用s，现给出每周生产单位酸奶的价格和每周要向客户提供的量，求最小花费。

解题思路:维护一个最小价格，每周都使用这个最小价格进行供奶。

min_cost =  min(min_cost + s, list[i].cost)

AC代码

#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
struct node {
	int cost, need;
}list[10001];
int main() {
	int n, s; long long ans;
	while (cin >> n >> s) {
		for (int i = 1; i <= n; i++) {
			scanf("%d%d", &list[i].cost, &list[i].need);
		}
		int min_cost = list[1].cost; ans = (long long)(list[1].cost * list[1].need);
		for (int i = 2; i <= n; i++) {
			if (min_cost + s < list[i].cost) {
				min_cost += s;
			}
			else { min_cost = list[i].cost;}
			ans += (long long)(min_cost * list[i].need);
		}
		cout << ans << endl;
	}
}