poj 3320 Jessica's Reading Problem(尺取法+map)_poj 3320 jessica's reading problem(尺取法)-优快云博客

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Jessica's Reading Problem

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13208		Accepted: 4544

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

5
1 8 8 8 1

Sample Output

一道水题，唯一的难点是如何保存每一段区间内不同数字出现的个数

用简单的visit数组，题目中数字可以达到2^32，数组开不下，所以就要用map容器来形成映射，使数字和数字在visit数组中的下标形成对应

当时太傻了，连map都想不出来。。。。。。

算法就是简单的尺取法模板

#include<stdio.h>
#include<string.h>
#include<map>
#define MAXN 1000010
#define INF 1999999999
using namespace std;

int book[MAXN];
map<int,int> fv;
int visit[MAXN];

int main()
{
    int p,s;
    scanf("%d",&p);
    s=0;
    for(int i=0;i<p;i++)
    {
        scanf("%d",&book[i]);
        if(fv.count(book[i])==0)
        {
            fv[book[i]]=s;
            s++;
        }
        visit[fv[book[i]]]++;

    }
    memset(visit,0,sizeof(visit));

    int l,r,sum,ans;
    l=r=sum=0;
    ans=INF;
    while(1)
    {
        while(sum<s&&r<p)
        {
            if(visit[fv[book[r]]]==0)
            {
                sum++;
            }
            visit[fv[book[r]]]++;
            r++;
        }
        if(sum<s) break;

        if(r-l<ans) ans=r-l;
        visit[fv[book[l]]]--;
        if(visit[fv[book[l]]]==0)
            sum--;
        l++;
    }
    printf("%d\n",ans);
    return 0;

}

算法就是简单的尺取法