P3805 【模板】manacher

合集 - 字符串(13)

1.P8736 [蓝桥杯 2020 国 B] 游园安排2024-03-302.P4341 [BJWC2010] 外星联络2024-04-013.P4391 [BOI2009] Radio Transmission 无线传输2024-04-044.P3435 [POI2006] OKR-Periods of Words2024-04-045.G1. Division + LCP (easy version)2024-05-046.P1435 [IOI2000] 回文字串2024-05-087.P4290 [HAOI2008] 玩具取名2024-05-138.F. 十六进制的异或2024-05-149.D. "a" String Problem2024-06-1110.B. Camp Schedule2024-07-18

11.P3805 【模板】manacher2024-07-20

12.P4824 [USACO15FEB] Censoring S2024-07-2113.P4555 [国家集训队] 最长双回文串2024-07-22

收起

原题链接

题解

细节

所有字符的回文半径初始化为 1
rmax=1
ans=1

code

#include<bits/stdc++.h>
#define ll long long
using namespace std;


void solve()
{
    string s;
    cin>>s;

    string s1;
    for(int i=0;s[i];i++)
    {
        s1+='#';
        s1+=s[i];
    }
    s1+='#';

    ll n=s1.size();
    vector<ll> r(n+5,1);
    ll rmax=0;//和自己匹配肯定是回文串
    ll ans=1,center=0;

    for(int i=0;i<n;i++)
    {
        if(i<rmax) r[i]=min(rmax-i,r[2LL*center-i]);

        while( i-r[i]>=0 && i+r[i]<n &&s1[i+r[i]]==s1[i-r[i]] ) r[i]++;

        if(i+r[i]>rmax)
        {
            center=i;
            rmax=i+r[i];
        }

        if(s1[i]=='#')
        {
            ans=max(ans,r[i]-1);
        }
        else
        {
            ans=max(ans,(r[i]-1)/2*2LL+1);
        }
    }

    cout<<ans<<'\n';
}
int main()
{
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int t=1;
    //cin>>t;
    while(t--) solve();
    return 0;
}