【二分】Jessica's Reading Problem

[POJ3320]Jessica's Reading Problem

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13095		Accepted: 4495

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

 

题目大意：选一段长度为k的区间使其包含所有元素，询问最小k

试题分析：二分区间长度，滑动窗口(尺取法)，离散化

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
inline long long read(){
    long long x=0,f=1;char c=getchar();
    for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
    for(;isdigit(c);c=getchar()) x=x*10+c-'0';
    return x*f;
}
#define LL long long
const int MAXN=1000001;
const int INF=999999;
const int mod=999993;
int N;
long long a[1000001];
int Hash[1000001];
long long A[1000001];
int tmp;

bool check(int k){
    memset(Hash,0,sizeof(Hash));
    int sum=0;
    for(int i=1;i<=k;i++){
        if(!Hash[A[i]]) sum++;
        Hash[A[i]]++;
    }
    if(sum==tmp) return true;
    for(int i=2;i+k-1<=N;i++){
        Hash[A[i-1]]--;
        if(!Hash[A[i-1]]) sum--;
        if(!Hash[A[i+k-1]]) sum++;
        Hash[A[i+k-1]]++;
        if(sum==tmp) return true;
    }
    return false;
}

int main(){
    N=read();
    for(int i=1;i<=N;i++){
        A[i]=a[i]=read();
    }
    sort(a+1,a+N+1);
    for(int i=1;i<=N;i++) A[i]=lower_bound(a+1,a+N+1,A[i])-a;
    for(int i=1;i<=N;i++) if(!Hash[A[i]]) Hash[A[i]]++,tmp++;
    int l=1,r=N;
    while(l<=r){
        int mid=(l+r)>>1;
        if(check(mid)) r=mid-1;
        else l=mid+1;
    }
    printf("%d
",l);
}