poj 3320 jessica's Reading PJroblem 尺取法 -map和set的使用

jessica's Reading PJroblem

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9134		Accepted: 2951

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：给你n个数字序列，求最小的子序列的长度，包含所有种类的数字

分析： 尺取法，能用尺取法的原因：假设当前a[s],a[s+1],,,,,a[t]包含只有种类，当s=s+1时，要仍然包含所有种类，则t'>=t，

        关键是set和map的使用 ，set相当于集合,map相当于数组，map数组可以开的大小范围不知道。。

int num[1000100],a[1000100];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        set<int > w;
        memset(num,0,sizeof(num));
        for(int i=1;i<=n;i++)
           {
               scanf("%d",&a[i]);
               w.insert(a[i]);
           }
        int cnt=w.size();
        int p=1,q=1,t=1,res=n+1;num[a[p]]++;//第一次re的代码，因为a[i]是整形范围的
//所以num就会爆内存，故只能换map了

　　下面是AC代码

#include<cstdio>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include<map>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long LL;
typedef unsigned long long ULL;
const int mod = 1000000007;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f;
int a[1000100];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        set<int> w;map<int,int> num;
        for(int i=1;i<=n;i++)
           {
               scanf("%d",&a[i]);
               w.insert(a[i]);//set相当于集合
           }  
        int cnt=w.size();   
        int p=1,q=1,t=1,res=n+1;num[a[p]]++;
        for(;;)
        {
            while(t<cnt&&q<=n){
                  q++;
                  if(!num[a[q]])
                    t++;
                  num[a[q]]++;
            }
            if(t<cnt)
                break;
            if(res>q-p+1) res=q-p+1;
            num[a[p]]--;
            if(!num[a[p]]) t--;
            p++;
        }
        printf("%d
",res);
    }
    return 0;
}