POJ2393:Yogurt factory

2393:Yogurt factory

总时间限制: 

1000ms

 

内存限制: 

65536kB

描述

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.

输入

* 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.

输出

* 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.

样例输入

4 5
88 200
89 400
97 300
91 500

样例输出

126900

提示

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】 

有一个制作酸奶的工厂，需要在接下来的N(1<=N<=10000)周里
生产酸奶，第i周的成本为Ci(1<=Ci<=5000)，工厂有一个无限大的仓库可
以存储生产好了的酸奶，并且这些酸奶不会过期，可以一直存储，但是存
储每单位酸奶每周要花费S(1<=S<=100)。计算：要满足条件：工厂第i周必
须送出Yi(1<=Yi<=10000)单位的酸奶(可以从仓库中送出，也可以本周生产)，
最小花费是多少？

【样例说明】：第一周生产200单位酸奶并且全部送出；第二周，生产700单位：送出400，存储300； 第三周，送出第二周存储的300单位； 第四周，生产500单位并全部送出。

 

【题目分析】 第i周的酸奶可以是任意小于i的周生产的。于是如果第i周的酸奶在第j(i>j)周生产，相当于这酸奶在第j周生产，并且花费是第i周的成本加上(i-j)周的储存费用。

第i周的最小花费是可以由第i-1周的最小花费推出的：mincost(i) = min( mincost(i-1)+s,Ci );而第一周的最小成本就是C1，因为第一周的酸奶只能第一周生产。

于是从第一周开始可以逐步推出所有周的最小花费：

　　

for( int i=1; i<=N-1; i++ ) {
week[i].C = min( week[i-1].C+S,week[i].C );
//i推i+1,到N-1时就已经推出N了 N单独处理就行了
}

 

关于数据范围：最坏的情况是，有10000周，每周都需要送出10000单位酸奶，
并且每周的成本都是5000.在这种情况下，共需要10000*5000*10000=5e+11的花费。
所以使用long long 就可以了。

 

【代码】

#include <iostream>
#include <algorithm>
using namespace std;

const int maxn = 10010;

struct _week
{
    int C;
    int Y;
};
_week week[maxn];
int N,S;

int main()
{
    cin >> N >> S;
    long long totalcost = 0;
    for( int i=1; i<=N; i++ ) {
        cin >> week[i].C >> week[i].Y;
    }

    //推出2到N-1周的最小花费同时计算总费用
    for( int i=1; i<=N-1; i++ ) {
        totalcost += week[i].C * week[i].Y;
        week[i+1].C = min( week[i].C+S,week[i+1].C );
    }

    //总费用加上最后一周(第N周)的花费
    totalcost += week[N].C * week[N].Y;
    cout << totalcost << endl;
    return 0;
}

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