ACM常用基础——尺取法
尺取法：简单点来理解，就是在对一组数据处理的时候，每次选取适当的标尺(标尺左端，右端分别对应数组的下标i,j)；在满足题目条件之间，不断往其中加入数据(标尺内容+=a[j] ; j++) ，直到满足所给条件。这时固定右端 j 不动，右移左端 i ，还是满足条件的话，继续右移左端 i ，直到满足所给条件；之后继续移动左端 i,如此循环往复，直到右端 j 到达数组末尾；
简单例子：
代码：
#include
#include
using namespace std;
int n;
int a[200];
int main()
{
int n,max,sum;
cin>>n>>max;
for(int k=0;k>a[k];
int i=0,j=0,sum=0,ans=n+1;
while(1)
{
while(j=n且sum小于max则输出
ans=min(j-i,ans); //记录最小的间距
sum-=a[i++]; //i向右移动
}
cout< 例题：Codeforces Round #364 div.2 C. They Are Everywhere
Sergei B., the young coach of Pokemons, has found the big house which consists of nflats ordered in a row from left to right. It is possible to enter each flat from the street. It is possible to go out from each flat. Also, each flat is connected with the flat to the left and the flat to the right. Flat number 1 is only connected with the flat number 2 and the flat number n is only connected with the flat number n - 1.
There is exactly one Pokemon of some type in each of these flats. Sergei B. asked residents of the house to let him enter their flats in order to catch Pokemons. After consulting the residents of the house decided to let Sergei B. enter one flat from the street, visit several flats and then go out from some flat. But they won't let him visit the same flat more than once.
Sergei B. was very pleased, and now he wants to visit as few flats as possible in order to collect Pokemons of all types that appear in this house. Your task is to help him and determine this minimum number of flats he has to visit.
Input
The first line contains the integer n (1 ≤ n ≤ 100 000) — the number of flats in the house.
The second line contains the row s with the length n, it consists of uppercase and lowercase letters of English alphabet, the i-th letter equals the type of Pokemon, which is in the flat number i.
Output
Print the minimum number of flats which Sergei B. should visit in order to catch Pokemons of all types which there are in the house.
Examples
Input
3
AaA
Output
2
Input
7
bcAAcbc
Output
3
Input
6
aaBCCe
Output
5
Note
In the first test Sergei B. can begin, for example, from the flat number 1 and end in the flat number 2.
In the second test Sergei B. can begin, for example, from the flat number 4 and end in the flat number 6.
In the third test Sergei B. must begin from the flat number 2 and end in the flat number 6.
题目大意：一共n个房子，里面有n个神奇宝贝（有重复的），求 存在所有种类神奇宝贝 的最小区间；
如样例3： 6 aaBCCe 所求区间应该是 aBCCe 所以结果是5；（想想其实和求最小区间和的题意思一样）
我的解法：1 :用set先求出总的种类数，保存到int max;
2 :用一个map
3 :往里面加元素，就popc[元素]++；删除即popc[元素]--；这里要注意map的特性：当它里面元素的个数减为0的时候，储存这个数据的区域还在；即这时候map的size不会自动减一：如图：
以下是两种风格的尺取做法，请自行对比阅读：
代码1：
#include
#include
#include
#include 代码2：
#include
#include
#include
#include