poj-3320-Jessica's Reading Problem



                                                Jessica's Reading Problem

Time Limit: 1000MSMemory Limit: 65536KTotal Submissions: 9968Accepted: 3307

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 containsP 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

51 8 8 8 1

Sample Output

2

Source

POJ Monthly--2007.08.05, Jerry

题意：告诉你一个n页的书，以及每一页上的知识点类型，要你求某个区间的长度，使得这个区间包含整本书的知识点，并且总页数最少。

题目链接：Jessica's Reading Problem

解题思路：因为要求区间是连续的，用一个集合记住这个区间的知识点是否出现过，在第一次找到包含所有知识点时，因为要使区间最短，可以逐步推进起点，直至推进起点后的区间不再包含所有在知识点，继续推进终点。

对于集合，可以用set来存储，知识点出现的次数用map来维护，再一次见证了STL的方便性！

代码：