POJ 3320 Jessica's Reading Problem 尺取法 - 细语呢喃 - 博客频道 - CSDN.NET

http://poj.org/problem?id=3320

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.

题目大意：
某人读一本书，要看完所有的知识点，这本书共有P页，第i页恰好有一个知识点ai，（每一个知识点都有一个整数编号）。全书同一个知识点可能会被提到多次，他希望阅读其中一些连续的页把所有知识点都读到，给定每页所读到的知识点，求最少的阅读页数。
思路：
和上一题一样，也是尺取法的应用。
假设从某一页s开始阅读，为了覆盖所有的知识点读到t页，这样的话如果从s+1开始阅读，那么必须读到t'>=t位置，故可以用尺取法。
用上Map来统计次数，取出前一项要把对应的知识点的编号次数-1.详见代码。

1. int a[MAXN];  
2. set<int> s;  
3. map<int,int> x;  
4. int main()  
5. {  
6.     int p;  
7.     while(~scanf("%d",&p))  
8.     {  
9.         for(int i=0;i<p;i++)  
10.             scanf("%d",&a[i]);  
11.           
12.         s.clear();  
13.         for(int i=0;i<p;i++)  
14.             s.insert(a[i]);  
15.         int n=s.size();  
16.         int s=0,t=0,ans=INF,cnt=0;  
17.         while(true)  
18.         {  
19.             while(cnt<n && t<p)  
20.             {  
21.                 if(x[ a[t++] ]++==0)  
22.                     cnt++;  
23.             }  
24.             if(cnt<n)  break;  
25.             ans=min(ans,t-s);  
26.             if(--x[a[s++]]==0)  
27.                 cnt--;        
28.         }  
29.         printf("%d\n",ans);  
30.     }  
31.     return 0;  
32. } 

Read full article from POJ 3320 Jessica's Reading Problem 尺取法 - 细语呢喃 - 博客频道 - CSDN.NET