POJ3320 Jessica's Reading Problem【尺取法】

Jessica's Reading Problem

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12441		Accepted: 4234

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

问题链接：POJ3320 Jessica's Reading Problem。

题意简述：一本书有P页，每一页都一个知识点，求最少的连续页数覆盖所有的知识点。

问题分析：

用尺取法来解决。每读一页知识点就有可能增加，是单调递增的，满足用尺取法解决问题的条件。

先求出最前面的子序列，使之覆盖所有知识点；

重复后面一步，直到p个元素都使用到；

去掉子序列的第1个元素，子序列的后面再加上其他元素，使之满足子序列覆盖所有知识点的条件，找出最小的长度。

程序说明：（略）

AC的C++语言程序如下：

/* POJ3320 Jessica's Reading Problem */

#include <iostream>
#include <set>
#include <map>
#include <stdio.h>

using namespace std;

const int N = 1000000;
int a[N+1];

int solve(int p)
{
    // 统计总共有多少个知识点，存储于变量n中
    set<int> eset;
    for(int i=0; i<p; i++)
        eset.insert(a[i]);
    int n = eset.size();

    // 使用map维护尺取法过程中的知识点的数量的增减
    map<int, int> count;
    int start=0, end=0, num=0, res=p;
    for(;;) {
        while(end < p && num < n) {
            if(count[a[end]]++ == 0)
                num++;
            end++;
        }
        if(num < n)
            break;
        res = min(res, end - start);
        if(--count[a[start++]] == 0)
            num--;
    }
    return res;
}

int main()
{
    int p;

    while(scanf("%d", &p) != EOF) {
        for(int i=0; i<p; i++)
            scanf("%d", &a[i]);

        printf("%d\n", solve(p));
    }

    return 0;
}