poj 3320 Jessica's Reading Problem（尺取法,map，set）

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Jessica面临期末考试，需从一本厚书中掌握所有知识点。为高效复习，她决定仅阅读包含全部知识点的最短连续部分。本题需找出这部分的长度。

Jessica's Reading Problem

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10071		Accepted: 3352

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

题意：题意：每一页有一个知识点，问学习所有知识点至少需要看几页

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <set>
#include <map>
using namespace std;
#define N 1000010
int a[N];
map<int,int>mp;
set<int>num;
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        mp.clear();
        num.clear();
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            num.insert(a[i]);
        }
        int m=num.size();
        int l=0,r=0;
        int cnt=0;
        int ans=n;
        while(l<n)
        {
            while(cnt<m&&r<n)
            {
                if(mp[a[r]]==0) cnt++;
                mp[a[r]]++;
                r++;
            }
            if(cnt<m) break;
            ans=min(ans,r-l);
            mp[a[l]]--;
            if(!mp[a[l]])
                cnt--;
            l++;
        }
        printf("%d\n",ans);
    }
    return 0;
}