假设没有操作1,就是裸的回文串自己主动机......

能够从头部插入字符的回文串自己主动机,维护两个last点就好了.....

当整个串都是回文串的时候把两个last统一一下

Victor and String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/262144 K (Java/Others)
Total Submission(s): 30    Accepted Submission(s): 9

/* ***********************************************
Author        :CKboss
Created Time  :2015年08月24日 星期一 10时32分04秒
File Name     :HDOJ5421.cpp
************************************************ */#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>using namespace std;typedef long long int LL;const int maxn=200100;
const int C=30;int L,R;
int nxt[maxn][C];
int fail[maxn];
LL cnt[maxn];
LL num[maxn];
int len[maxn];
int s[maxn];
int last[2],p,n;
LL tot;int newnode(int x)
{for(int i=0;i<C;i++) nxt[p][i]=0;num[p]=0; len[p]=x;return p++;
}void init()
{p=0;newnode(0); newnode(-1);memset(s,-1,sizeof(s));L=maxn/2; R=maxn/2-1;last[0]=last[1]=0;fail[0]=1;tot=0;
}/// d=0 add preffix d=1 add suffixint getfail(int d,int x)
{if(d==0) while(s[L+len[x]+1]!=s[L]) x=fail[x];else if(d==1) while(s[R-len[x]-1]!=s[R]) x=fail[x];return x;
}void add(int d,int c)
{c-='a';if(d==0) s[--L]=c;else s[++R]=c;int cur=getfail(d,last[d]);if(!nxt[cur][c]){int now=newnode(len[cur]+2);fail[now]=nxt[getfail(d,fail[cur])][c];nxt[cur][c]=now;num[now]=num[fail[now]]+1;}last[d]=nxt[cur][c];if(len[last[d]]==R-L+1) last[d^1]=last[d];tot+=num[last[d]];
}int main()
{//freopen("in.txt","r",stdin);//freopen("out.txt","w",stdout);int _;while(scanf("%d",&_)!=EOF){init();while(_--){int kind;char ch[5];scanf("%d",&kind);if(kind<=2){kind--;scanf("%s",ch);add(kind,ch[0]);}else if(kind==3) printf("%d\n",p-2);else if(kind==4) printf("%lld\n",tot);}}return 0;
}