3280 Cheapest Palindrome(尺取法)

本文介绍了一种高效方法帮助Jessica通过阅读最小数量的连续页面来覆盖整个教科书中涵盖的所有知识点,利用尺取法实现最优解。

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http://poj.org/problem?id=3320

                Jessica's Reading Problem

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


某人读一本书,要看完所有的知识点,这本书共有P页,第i页恰好有一个知识点ai,(每一个知识点都有一个整数编号)。全书同一个知识点可能会被提到多次,他希望阅读其中一些连续的页把所有知识点都读到,给定每页所读到的知识点,求最少的阅读页数。

用到尺取法。

尺取的思路:

①不停扩展t,并把扫过知识点丢到map里,直到map的size符合要求。

②更新结果。

②s++,map里的对应mm(a[l++])的个数-1,相当于移出这页。

如果对应的mm的个数<=0,则应该erase掉这个mm,防止map::size()的误判。


#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
#include <set>
#include <algorithm>

using namespace std;

const int N = 1000000 + 100 ;
int arr[N];
map<int,int>mm;

int main()
{
    int n;
    scanf("%d",&n);

    for(int i=1;i<=n;i++) scanf("%d",&arr[i]),mm[arr[i]]++;

    int ans = n;
    int tmp = mm.size();
    int s = 1, t = 1;
    int num = 0;
    mm.clear();
    while(true)
    {
        while(t<=n && num<tmp) 
        {
           // if(mm[arr[t++]]++==0)  num++;
           if(mm[arr[t]]==0) num++;
           mm[arr[t]]++;
           t++;
        }

        if(num<tmp) break;
        ans = min(ans,t-s);

        //if(--mm[arr[s++]]==0) num--;
        mm[arr[s]]--;
        if(mm[arr[s]]==0) num--; 
        s++;
    }
    printf("%d\n",ans);
    return 0;
}
### Python 中获取最大值和最小值的方法 在 Python 中,`max()` 和 `min()` 是内置函数,用于分别找到组可比较元素中的最大值和最小值。这些函数可以应用于多种数据结构,例如列表、元组、字典以及自定义对象。 以下是具体的使用示例: #### 使用 `max()` 找到最大值 对于一个简单的列表或元组,可以直接调用 `max()` 函数来返回其中的最大值[^1]。 ```python numbers = [3, 5, 7, 2, 8] maximum_value = max(numbers) print(f"The maximum value is {maximum_value}") ``` 上述代码会输出: ``` The maximum value is 8 ``` #### 使用 `min()` 找到最小值 同样地,可以通过 `min()` 函数找到序列中的最小值[^2]。 ```python numbers = (3, 5, 7, 2, 8) minimum_value = min(numbers) print(f"The minimum value is {minimum_value}") ``` 运行结果为: ``` The minimum value is 2 ``` #### 应用于其他数据类型 除了基本的数据结构外,还可以通过指定参数的方式应用到更复杂的情况中。例如,在字典中查找键或值的最大/最小值[^4]。 ```python prices = {'apple': 0.99, 'banana': 0.59, 'cherry': 1.29} # 查找价格最低的商品名称 cheapest_item = min(prices, key=prices.get) print(f"The cheapest item is '{cheapest_item}' with price ${prices[cheapest_item]}") # 查找价格最高的商品名称 most_expensive_item = max(prices, key=prices.get) print(f"The most expensive item is '{most_expensive_item}' with price ${prices[most_expensive_item]}") ``` 执行以上脚本将得到如下输出: ``` The cheapest item is 'banana' with price $0.59 The most expensive item is 'cherry' with price $1.29 ``` #### 整型范围的特殊处理 如果目标是从理论上了解 Python 的整数边界,则需要注意 Python 并未像 Java 那样提供固定的上下限常量(如 `Integer.MIN_VALUE` 或 `Integer.MAX_VALUE`)。由于 Python 支持任意精度整数,因此实际上不存在硬编码的最大或最小值限制[^3]。不过,某些实现可能因内存或其他资源约束而有所差异。 ---
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