poj3320：Jessica's Reading Problem

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

题目大意：
为了准备考试，Jessica开始读一本很厚的课本。要想通过考试，必须把课本中所有的知识点都掌握。这本书总共有P页，第i页恰好有一个知识点ai（每个知识点都有一个整数编号）。全书中同一个知识点可能会被多次提到，所以她希望通过阅读其中一些连续的一些页把所有的知识点都覆盖到。给定每页写到的知识点，求出要阅读的最少页数。

解题思路
利用尺取法，在某个区间【s,t】已经覆盖所有知识点的情况下，判断【s+1,t’】(t’>t)。时间复杂度为O（PlogP）。

代码如下：

#include<iostream>#include<cstdio>#include<algorithm>#include<memory.h>#include<map>//知识点-出现次数的映射#include<set>//计算知识点的总个数using namespace std;int a[1000005];int P;void solve();int main(){    int i;    while(~scanf("%d",&P))    {        for(i=0;i<P;i++)        scanf("%d",&a[i]);        solve();    }    return 0;}void solve(){    set<int> all;    for(int i=0;i<P;i++)    all.insert(a[i]);    int n=all.size();    int s=0,t=0,num=0;    map<int,int> count;    int res=P;    for(;;)    {        while(t<P&&num<n)        {            if(count[a[t++]]++==0)//出现新的知识点            num++;        }        if(num<n)        break;        res=min(res,t-s);        if(--count[a[s++]]==0)//某个知识点的出现次数为0        num--;    }    printf("%d\n",res);}