【0】做题心得 - 2025 NOIP #67 - T1【贪心】

ExFish · 2025-11-24 10:16:02 · 题解

哦不其实是直接左右缩着然后直接最优化区间即可。然后注意要固定左右侧都要跑一遍。

#include<bits/stdc++.h>
using namespace std;
const int N=1e6+10,V=26+5;
int n,ans;
char a[N];
stack<int>stk[V];
int solve(){
    for(int c=0;c<26;c++)
        while(stk[c].size()) stk[c].pop();
    for(int i=1;i<=n;i++)
        stk[a[i]-'a'].push(i);
    int L=n,R=0,res=0;
    for(int i=1;i<=n&&L>i;i++){
        int ch=stk[a[i]-'a'].top(); stk[a[i]-'a'].pop();
        if(L<=ch&&ch<=R) continue;
        if(ch==n-i+1) res+=max(R-L+1,0), R=n-i;
        else L=min(L,ch), R=max(R,n-i+1); 
    }
    res+=max(R-L+1,0);
    return res;
}
int main(){
    freopen("poem.in","r",stdin);
    freopen("poem.out","w",stdout);
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin>>n;
    for(int i=1;i<=n;i++)
        cin>>a[i];
    ans=min((int)1e9,solve());
    reverse(a+1,a+n+1);
    ans=min(ans,solve());
    cout<<min(ans,n-1);
    return 0;
}