LINK
直接建回文树,得到以 i i i结尾/开始的最大回文串长度
然后枚举分割点 O ( n ) O(n) O(n)取 m a x max max即可
比较模板的题
#include <bits/stdc++.h>
using namespace std;
const int maxn = 4e5+10;
struct PAM
{
int fail[maxn],zi[maxn][27],len[maxn],id,las;
int pre[maxn],suf[maxn],qui[maxn];//[1,i]结束
char a[maxn];
PAM()
{
fail[0] = 1, fail[1] = 1; len[0] = 0, len[1] = -1;
las = id = 1;
}
int get_fail(int u,int index)
{
while( a[index]!=a[index-1-len[u]] ) u = fail[u];
return u;
}
void insert(int c,int index)
{
int u = get_fail(las,index);
if( zi[u][c]==0 )
{
int newnode = ++id, v = get_fail( fail[u],index);
len[newnode] = len[u]+2;
pre[index] = max( pre[index],len[newnode] );
suf[index-len[newnode]+1] = max( suf[index-len[newnode]+1],len[newnode] );
fail[newnode] = zi[v][c], zi[u][c] = newnode;
}
las = zi[u][c];
}
void solve()
{
cin >> ( a+1 ); int n = strlen( a+1 );
for(int i=1;i<=n;i++) insert( a[i]-'a',i );
int ans = 0;
for(int i=1;i<=n;i++) pre[i] = max( pre[i],1 ),suf[i] = max( suf[i],1 );
for(int i=1;i<=n;i++) pre[i] = max( pre[i],pre[i-1] );
for(int i=n;i>=1;i--) suf[i] = max( suf[i],suf[i+1] );
for(int i=2;i<=n;i++) ans = max( ans,pre[i-1]+suf[i] );
cout << ans;
}
}pam;
int main()
{
pam.solve();
}
