POJ - 3320 ：Jessica's Reading Problem（尺取 - STL(set,map)）

原题链接
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

题目描述

有n页，每页都有一个知识点的编号，要求得到连续的且含有所有知识点的页的最短长度
思路：说到连续、最短，我们很容易就想到了尺取，首先我们要得到总共有多少个知识点，然后对这些知识点编号进行尺取，如果某个区间中知识点数量等于全书的知识点数量，那么就记录并比较，然后去掉这个区间最后一个元素，（注意：这块需要判断，如果最后一个元素去掉之后该编号的知识点数变为0，那么计数知识点的变量-1），重复上面操作，就可以得到最短长度
代码如下

#include 
#include 
#include 
#include 
#include 

using namespace std;

set<int> s;
map<int,int> m;

int a[1000010];

int main()
{
    int i,j,p,n,num,ans;
    scanf("%d",&p);
    for(i = 0;i < p;i++)
    {
        scanf("%d",&a[i]);
  		//这里我们使用set容器，可以容易的知道总的知识点数
        s.insert(a[i]);
    }
    n = s.size();
    i = 0, j = 0, num = 0, ans = p;
    while(1)
    {
        while(j < p && num < n)
        {
            //使用map容器，很容易知道每个知识点该区间对应的数量
            if(m[a[j]] == 0)
                num++;
            m[a[j]]++;
            j++;
        }
        //当遍历该区间都没有知识点全部都存在，那么就不需要进行循环
        if(num < n)
            break;
        ans = min(ans, j - i);
        m[a[i]]--;
        //如果最后一个元素去掉之后该编号的知识点数变为0，那么num--
        if(m[a[i]] == 0)
            num--;
        i++;
    }
    printf("%d\n",ans);
    return 0;
}