POJ 3320 Jessica's Reading Problem（尺取法+STL）
Jessica's Reading Problem
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 18206 | Accepted: 6290 |
Description
Jessica's a very lovely girl wooed by lots of boys. Recently she has a problem. The final exam is coming, yet she has spent little time on it. If she wants to pass it, she has to master all ideas included in a very thick text book. The author of that text book, like other authors, is extremely fussy about the ideas, thus some ideas are covered more than once. Jessica think if she managed to read each idea at least once, she can pass the exam. She decides to read only one contiguous part of the book which contains all ideas covered by the entire book. And of course, the sub-book should be as thin as possible.
A very hard-working boy had manually indexed for her each page of Jessica's text-book with what idea each page is about and thus made a big progress for his courtship. Here you come in to save your skin: given the index, help Jessica decide which contiguous part she should read. For convenience, each idea has been coded with an ID, which is a non-negative integer.
Input
The first line of input is an integer P (1 ≤ P ≤ 1000000), which is the number of pages of Jessica's text-book. The second line contains P non-negative integers describing what idea each page is about. The first integer is what the first page is about, the second integer is what the second page is about, and so on. You may assume all integers that appear can fit well in the signed 32-bit integer type.
Output
Output one line: the number of pages of the shortest contiguous part of the book which contains all ideals covered in the book.
Sample Input
5
1 8 8 8 1
Sample Output
2题意描述：为了准备考试，Jessica开始读一本很厚的课本。要想通过考试，必须要课本上每个知识点看一遍。这本书一共有P页，第i页恰好有一个知识点a[i]，同一个知识点可能被多页反复提到。所以她想阅读连续的一些页把书本中所有的知识点全覆盖到。给定每页写到的知识点，求出阅读的最少页数。
分析：用stl里的set求出书中知识点的总个数，再用map对应每个知识点出现的次数，剩下的就跟尺取法的思路一样啦。
这里简单的总结下set用法：
set
set
set
set的各常见函数列表如下:
c++ stl容器set成员函数:begin()--返回指向第一个元素的迭代器
c++ stl容器set成员函数:clear()--清除所有元素
c++ stl容器set成员函数:count()--返回某个值元素的个数
c++ stl容器set成员函数:empty()--如果集合为空，返回true
c++ stl容器set成员函数:end()--返回指向最后一个元素的迭代器
c++ stl容器set成员函数:find()--返回一个指向被查找到元素的迭代器
c++ stl容器set成员函数:insert()--在集合中插入元素
c++ stl容器set成员函数:max_size()--返回集合能容纳的元素的最大限值
c++ stl容器set成员函数:size()--集合中元素的数目
代码如下：
#include
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