Poj 3320(two pointers)

Jessica's Reading Problem

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5967		Accepted: 1794

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

题意就是给一个数列，找到一个最小的区间，包含这个数列中出现的所有元素。这类连续的区间问题，很多都可以用这种尺取法线性的扫描。所谓的“尺取法”，就是维护两个指针，指向当前维护的区间的首和尾。维护是贪心的去尽量的缩小区间范围。比如这题就是要保证区间的首元素只出现了一次，否则就把首指针后移。当区间内不同元素的个数小于总个数，就把尾指针后移。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#include<iostream>
#include<algorithm>
#include<set>
#include<map>
typedef long long LL;
using namespace std;
const int maxn = 1000000 + 5;

map<int,int> M;
set<int> S;
int a[maxn];

int main(){
    int p;
    while(scanf("%d",&p) != EOF){
        S.clear();
        M.clear();
        for(int i = 0;i < p;i++){
            scanf("%d",&a[i]);
            S.insert(a[i]);
        }
        int n = S.size();
        if(n == 1){
            printf("1\n");
            continue;
        }
        int ans = p;
        int s = 0,t = 0,sum = 0;
        while(s < p && t < p){
            if(M[a[t]] == 0) sum++;
            M[a[t]]++;
            while(M[a[s]] >= 2){
                M[a[s]]--;
                s++;
            }
            if(sum == n){
                ans = min(ans,t-s+1);
                if(M[a[s]] == 1) sum--;
                M[a[s]]--;
                s++;
            }
            t++;
        }
        printf("%d\n",ans);
    }
    return 0;
}