NOI 2393:Yogurt factory【2019北大推免E】_北京大学yogurt factory-优快云博客

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NOI 2393:Yogurt factory【2019北大推免E】

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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.

题解

子问题：dp[i]：完成week[:i]的所有demand需要的最少cost
- 证明最优子结构（假设问题的最优解不蕴含子问题的最优解，即子问题存在更优的解）：dp[i]是完成week[:i]的所有demand需要的最少cost，其中完成week[:i-1]的所有demand需要的cost部分，假设不是最优（即不满足最优子结构性质），即存在更低的完成week[:i-1]的所有demand需要的cost，则这个更优解和dp[i]中完成week[i]的demand部分可以组合成问题dp[i]的最优解，与最优性矛盾
更新：dp[i]=dp[i-1]+选择week[:i]中在week[i]时单价最低的week生产本周的所有demand
- week[j]在week[i]时单价：cost[j]+S*(i-j) （初始生产的单价，加上每周存储的单价）
- note：
  - dp[i]对于dp[:i-1]只需要知道dp[i-1]，因此可以省略数组使用变量dp
  - 更新对于当前week week[i]，week[:i]的单价：（考虑从week[i-1]来的关系）
    - cost[j]+S*(i-j) = cost[j]+S*(i-1-j) +S
      - 进一步发现，所有之前已经有单价的单价都是同步+S，因此进入新的一周，最低price一定是min((原先的最低price+S), 新周的price)
      - 进一步的，这样的话cost也不用数组存了
    - 看到可以用在week[i-1]的单价来获取week[i]的单价，避免乘法
初始：dp[0]：前面没有周，必须本周完成demand，因此dp[0]=c[0]*d
时间复杂度：对于week[i]常数操作，因此复杂度是O(n)，N<=10^4<1-9是可以的
数据范围：
- 全周代价C*Y*N=5000*10^4*104 = 5*10^11>2*109用int不行要用long long
- 单价：1 <= C_i <= 5,000，1 <= S <= 100，1 <= N <= 10,000，则最大单价5000+100*10000<10^9，可以用int，不过和demand乘法时需要是long long乘法，所以建议用long long

#define MAXN 10005
int main()
{
    int N,S;
    cin>>N>>S;
    long long d,c;//demand, cost
    cin>>c>>d;//N>=1, this must be safe
    long long dp=c*d;
    long long min_price=c;
    for(int i=1; i<N; ++i){
        cin>>c>>d;
        //update min_price
        min_price=min(c,min_price+S);
        //update dp
        dp += min_price*d;
    }
    cout<<dp<<endl;
    return 0;
}