poj3320 尺取法

 

如题：http://poj.org/problem?id=3320

 

Jessica's Reading Problem

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7512 Accepted: 2387

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

 

 

题目大意：给出P页书和每一页书上的知识点编号，问看连续的几页要求遍历到所有知识点，输出最小页书。

 

思路：处理这种连续区间，尺取法特别管用，通过对条件的判断对一个数组开始和结束区间不断的调整，最终求出那个区间。

首先统计出知识点的个数，然后开始遍历数组，先令头=尾 s=t=1，然后找到第一个满足条件的尾部，也就是所有知识点都找出，然后去调整头部，每次删除一个头部，如果不满足尾部继续调整，如果删除了还满足，继续删除。两个指针一个相当去全部向前遍历一遍数组，负责度O(n).

 

#include<iostream>
#include<cstdio>
#include<set>
#include<map>
using namespace std;

#define min(a,b)(a<b?a:b)
int a[1000005];

int main()
{
 int P;
 cin>>P;
 int i;
 set<int>all;
 for(i=1;i<=P;i++)
 {
  scanf("%d",&a[i]);
  all.insert(a[i]);
 }
 int n=all.size();
 map<int,int>count;  //编号和出现的数量
 int s,t;
 s=t=1;
 int res=P;
 int num=0;
 while(1)
 {
  while(t<=P&&num<n)
   if(count[a[t++]]++==0)
    num++;
  if(num<n)
   break;
  res=min(res,t-s);
  if(--count[a[s++]]==0)
   num--;
 }
 cout<<res<<endl;
}