poj 2393 Yogurt factory

                                                                                                  Yogurt factory

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9431		Accepted: 4801

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.

 

题意：酸奶加工厂生产酸奶并准备卖出去，共有N周，每周都要送出规定量的酸奶，且每周生产单位吨奶耗费都不同，记为C_i,每周都有要送走的牛奶量Y_i,若每周生产量过多，囤积会有囤积费，每周单位吨奶的存储费为S。尽量花费最少的费用，求最小费用。

思路：

从每一周的单位酸奶送出量需要耗费的资金方面来考虑，设P为每一周单位酸奶送出量需要耗费资金的最小值，（除第一周外）每一周的P取值为min{这周的C_i,上周的P+S}。

AC代码：

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
using namespace std;
int main() {
    int N, s;
    cin >> N >> s;
    long long cost=0;
    int p = 5000;//p每次都代表当前一周每单位吨奶需要消耗的钱,要让其尽量的小
    for (int i = 0;i < N;i++) {
        int c, y;
        cin >> c >> y;
        p = min(c,p+s);//p取这周的单位生产价格或者上周的生产价格加上储存费的最小值
        cost += p*y;
    }
    cout << cost << endl;
    return 0;
}