The front and back ends are inserted at the same time, and the prefix fail pointer is added. It is found that for each node, the prefix fail and the suffix fail are the same. Then update the situation directly. And inserting characters in front will not affect the fail of the original tree. Because the fail length of each node is smaller than itself.
And it should be noted that when inserting a node, if the length of its newly produced palindrome is equal to the length of the total string at this time, L_last, R_last must be updated at the same time
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
///len表示当前节点对应回文子串长度
///num表示前pam节点(即读入s[i]之后,s[1,…,i]的最长回文后缀对应节点u)对应回文子串的所有不同回文后缀的个数（包括自身）
///sz表示当前节点对应回文子串在主串中出现次数(需要处理完整个主串后倒着对failDP)
///别忘初始化！！！！
///内存问题
const int SZ = 26;///字符集
const int maxn = 2e5 + 6; ///开两倍
struct PAM {
struct PamNode{
int fail,trans[SZ],sz,len,num;}pam[maxn];
int tot,n,l,r;char s[maxn]; ll ans;
void init() {
tot=1; ans=0; l=100003,r=100002;
memset(s,-1,sizeof s); ///特别注意
pam[0].fail=1; pam[0].len=0;
pam[1].fail=1; pam[1].len=-1;
memset(pam[0].trans,0,SZ*sizeof (int));
memset(pam[1].trans,0,SZ*sizeof (int));
}
inline int newnode(int len) {
tot++;
memset(pam[tot].trans,0,SZ*sizeof (int));
pam[tot].len=len; pam[tot].fail=0;
pam[tot].sz=0; pam[tot].num=0;
return tot;
}
inline int L_getfail(int i,int u) {
while(s[i+pam[u].len+1]^s[i]) u=pam[u].fail;
return u;
}
inline int R_getfail(int i,int u) {
while(s[i-pam[u].len-1]^s[i]) u=pam[u].fail;
return u;
}
inline int L_append(char c,int u) {
s[--l]=c; c=c-'a';
int fa=L_getfail(l,u);
u=pam[fa].trans[c];
if(!u) {
int z=newnode(pam[fa].len+2);
int w=L_getfail(l,pam[fa].fail);
pam[z].fail=pam[w].trans[c];
u=pam[fa].trans[c]=z;///注意这里要后更新，否则上面getfail时可能导致死循环
pam[z].num=pam[pam[z].fail].num+1;
}
ans=ans+pam[u].num;
pam[u].sz++;return u;
}
inline int R_append(char c,int u) {
s[++r]=c; c=c-'a';
int fa=R_getfail(r,u);
u=pam[fa].trans[c];
if(!u) {
int z=newnode(pam[fa].len+2);
int w=R_getfail(r,pam[fa].fail);
pam[z].fail=pam[w].trans[c];
u=pam[fa].trans[c]=z;///注意这里要后更新，否则上面getfail时可能导致死循环
pam[z].num=pam[pam[z].fail].num+1;
}
ans=ans+pam[u].num;
pam[u].sz++;return u;
}
void calu() {
for(int i=tot,fail;i>1;i--) {
fail=pam[i].fail;
pam[fail].sz+=pam[i].sz;
}
}
}pa;
int n,k;
int main() {
while(~scanf("%d",&n)) {
pa.init(); char op[3];
for(int i=1,L_last=0,R_last=0;i<=n;i++) {
scanf("%d",&k);
if(k==1) {
scanf("%s",op);
L_last=pa.L_append(op[0],L_last);
if(pa.pam[L_last].len==pa.r-pa.l+1) R_last=L_last;
}
if(k==2) {
scanf("%s",op);
R_last=pa.R_append(op[0],R_last);
if(pa.pam[R_last].len==pa.r-pa.l+1) L_last=R_last;
}
if(k==3) printf("%d\n",pa.tot-1);
if(k==4) printf("%lld\n",pa.ans);
}
}
}