双倍回文
#include <bits/stdc++.h>
using namespace std;
const int M=1e6+5;
char s[M];
int p[M],n=1,ans=0;
void Manacher(string t) {
s[0]='@',s[1]='#';
for(auto i:t)s[++n]=i,s[++n]='#';
for(int i=1,mid=0,r=0;i<=n;i++) {
if(i<=r)p[i]=min(p[2*mid-i],r-i);
while(s[i-p[i]]==s[i+p[i]]) {
p[i]++;
if(i+p[i]-1>r&&s[i]=='#'&&(p[i]-1)%4==0) {
if(p[i-(p[i]-1)/2]-1>=(p[i]-1)/2)ans=max(ans,p[i]-1);
}
}
if(i+p[i]>r)r=i+p[i]-1,mid=i;
}
}
//实质上是枚举了每一个出现过的回文串的
//也就是对每一个回文串都判断一下
int main() {
int n;
string t;
cin>>n>>t;
Manacher(t);
cout<<ans<<'\n';
return 0;
}
标签:int,Manacher,双倍,mid,ans,回文
From: https://www.cnblogs.com/basicecho/p/17120750.html