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

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 numbern - 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

题目大意等价于求最小的区间包含所有种类的数.

尺取法。枚举左端点,右端点通过种类数决定

/* ***********************************************
Author        :guanjun
Created Time  :2016/8/8 14:04:46
File Name     :cf364c.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10010
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;
priority_queue<int,vector<int>,greater<int> >pq;
struct Node{
    int x,y;
};
struct cmp{
    bool operator()(Node a,Node b){
        if(a.x==b.x) return a.y> b.y;
        return a.x>b.x;
    }
};

bool cmp(int a,int b){
    return a>b;
}
map<char,int>mp;
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    int n,tot;
    string s;
    while(cin>>n){
        cin>>s;
        tot=0;
        mp.clear();
        for(int i=0;i<n;i++){
            if(!mp[s[i]])mp[s[i]]++,tot++;
        }
        mp.clear();
        int r=0,l=0;
        int ans=n,num=0;
        for(int l=0;l<n;l++){
            while(r<n&&num<tot){
                if(!mp[s[r++]]++)num++;
            }
            if(num<tot)break;
            ans=min(ans,r-l);
            if(--mp[s[l]]==0)num--;
        }
        cout<<ans<<endl;
    }
    return 0;
}
原文地址:https://www.cnblogs.com/pk28/p/5749517.html