TopCoder----卖柠檬

1. 题目描述

Problem Statement

You are playing a game called Slime Tycoon.You will be selling Slimonades in this game, and your goal is to sell as many as you can.

The game will consist of N game days, numbered 0 through N-1 in order.You are given two vector <int>s morning and customers with N elements each, and an int
stale_limit.These represent constraints on how many Slimonades you can produce and sell, as explained below.



In each game day, three things happen, in the following order:

• Early in the morning of day i: All Slimonades that were produced stale_limit days ago (i.e., on day i-stale_limit) go stale. You cannot sell stale Slimonades, you must throw them away immediately.
• During day i: You can produce at most morning[i] new Slimonades. (Formally, you choose an integer X between 0 and
  morning[i], inclusive, and produce X Slimonades.)
• In the evening of day i: You can sell at most customers[i] Slimonades. (That is, if you have at mostcustomers[i] Slimonades, you sell all of them. Otherwise, you sell exactly customers[i] Slimonades. In
  that case, you get to choose which Slimonades you sell and which ones you keep for later days.)

What is the maximum total number of Slimonades that you can sell during these N days?

Definition

Class:	SlimeXSlimonadeTycoon
Method:	sell
Parameters:	vector <int>, vector <int>, int
Returns:	int
Method signature:	int sell(vector <int> morning, vector <int> customers, int stale_limit)
(be sure your method is public)

Limits

Time limit (s):	2.000
Memory limit (MB):	256

Constraints

-

morning will contain between 2 and 50 elements, inclusive.

-

Each element of morning will be between 0 and 10000, inclusive.

-

customers will contain the same number of elements as
morning.

-

Each element of customers will be between 0 and 10000, inclusive.

-

stale_limit will be between 1 and N, inclusive.

Examples

{5, 1, 1}

{1, 2, 3}

2

Returns: 5

Here's one optimal solution.

• Day 0: We produce 4 Slimonades, then sell 1 of them.
• Day 1: We produce 1 Slimonade (so now we have 4). In the evening, we sell two of the Slimonades that were made yesterday.
• Day 2: We still have one Slimonade that was made on day 0. It goes stale and we throw it away. We produce one more Slimonade. In the evening, we sell 2 Slimonades (the one made yesterday and the one made today).

2. 分析

采用如下的策略，每天尽可能多地生产，并且优先卖生产日期久的。。采用一个队列。。

3. 代码

class SlimeXSlimonadeTycoon
{
public:
int sell(vector <int> morning, vector <int> customers, int stale_limit)
{
  if(morning.empty()) return 0;

  int que[100000];
  int front = 0, rear = 0;
  int count(0);
  for(int i=0; i<morning.size(); ++i)
  {
    if(front - rear == stale_limit) ++rear;
    que[front++] = morning[i];

    while(rear < front && customers[i] > 0)
    {
      if(que[rear] > customers[i]){
        count += customers[i];
        que[rear] -= customers[i];
        break;
      }
      else{
        count += que[rear];
        customers[i] -= que[rear];
        ++rear;
      }
    }
  }
return int(count) ;
}
};