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

51 8 8 8 1

Sample Output

2

Source

POJ Monthly--2007.08.05, Jerry

題意就是給一個數列，找到一個最小的區間，包含這個數列中出現的所有元素。這類連續的區間問題，很多都可以用這種尺取法線性掃描。所謂的“尺取法”，就是維護兩個指標，指向當前維護的區間的首和尾。維護是貪心的去盡量的縮小區間範圍。比如這題就是要保證區間的首元素只出現了一次，否則就把首指標後移。當區間內不同元素的個數小於總個數，就把尾指標後移。

#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;}