（POJ - 3320）Jessica's Reading Problem 尺取法 set map

POJ - 3320
Jessica’s Reading Problem
Time Limit: 1000MS Memory Limit: 65536KB 64bit IO Format: %lld & %llu
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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
Source
POJ Monthly–2007.08.05, Jerry

题意：
给你P个整数（可以有有重复），问你包含了所有不同数的最短区间的长度？

分析：、
假设某个区间[s,t]已经覆盖了所有的知识点，那么下一个区间[s+1,t’]如何求出？
所有的知识点都覆盖了==每个知识点出现的次数>=1
所以我们可以利用map来记录每个知识点出现的次数。这样把区间最开头的页s去除后便可以判断[s+1，t]是否满足条件。
从区间的开头把s取出之后，页s上出现的知识点出现的次数就减1，如果此时这个知识点的出现次数为0，在同一个知识点再次出现之前，不停的将区间末尾t向后推进即可。每次将区间末尾追加的知识点的出现的次数加1.