Codeforces Round #364 (Div. 2) C. They Are Everywhere （尺取法）

C. They Are Everywhere

time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Sergei B., the young coach of Pokemons, has found the big house which consists of n flats 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.

题意：给你一个字符串，然后求将字符串中出现的字符都包括的连续的子串的最短长度。

思路：标记字符，然后按照尺取法来扫一遍就可以了。

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<map>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010000
#define LL long long
#define ll __int64
#define INF 0x7fffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
#define eps 1e-8
using namespace std;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
double dpow(double a,ll b){double ans=1.0;while(b){if(b%2)ans=ans*a;a=a*a;b/=2;}return ans;}
//head
int v[3][100];
char s[MAXN];
int main()
{
    int len;scanf("%d",&len);
    scanf("%s",s);mem(v);
    int cnt=0;
    for(int i=0;i<len;i++)
    {
        if(s[i]>='A'&&s[i]<='Z')
        {
            if(v[1][s[i]-'A']==0) cnt++;
            v[1][s[i]-'A']++;
        }
        else if(s[i]>='a'&&s[i]<='z')
        {
            if(v[2][s[i]-'a']==0) cnt++;
            v[2][s[i]-'a']++;
        }
    }
    int r=0,l=0,num=0,ans=INF;mem(v);
    for(;;)
    {

        while(r<len&&num<cnt)
        {
            if(s[r]>='A'&&s[r]<='Z')
            {
                if(v[1][s[r]-'A']==0) num++;
                v[1][s[r]-'A']++;
            }
            else if(s[r]>='a'&&s[r]<='z')
            {
                if(v[2][s[r]-'a']==0) num++;
                v[2][s[r]-'a']++;
            }
            r++;
        }
        if(num<cnt)
        break;
        ans=min(ans,r-l);
        if(s[l]>='A'&&s[l]<='Z')
        {
            v[1][s[l]-'A']--;
            if(v[1][s[l]-'A']==0)
                num--;
        }
        else if(s[l]>='a'&&s[l]<='z')
        {
            v[2][s[l]-'a']--;
            if(v[2][s[l]-'a']==0)
                num--;
        }
        l++;
    }
    printf("%d\n",ans);
    return 0;
}