NC18386 字符串(尺取)

题目链接

题意：



题解：

AC代码

/*
    Author:zzugzx
    Lang:C++
    Blog:blog.csdn.net/qq_43756519
*/
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define endl '\n'
#define SZ(x) (int)x.size()
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int mod=1e9+7;
//const int mod=998244353;
const double eps = 1e-10;
const double pi=acos(-1.0);
const int maxn=1e6+10;
const ll inf=0x3f3f3f3f;
const int dir[][2]={{0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1}};

int cnt[26];
bool ok(){
    for(int i=0;i<26;i++)
        if(!cnt[i])return 0;
    return 1;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    string s;
    cin>>s;
    int n=s.length();s='.'+s;
    int l=1,ans=inf;
    for(int r=1;r<=n;r++){
        cnt[s[r]-'a']++;
        while(ok())ans=min(ans,r-l+1),cnt[s[l++]-'a']--;
    }
    cout<<ans;
    return 0;
}