HDOJ 5421 Victor and String 回文串自己主动机

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

Problem Description

Victor loves to play with string. He thinks a string is charming as the string is a palindromic string.

Victor wants to play n times. Each time he will do one of following four operations.

Operation 1 : add a char c to the beginning of the string.

Operation 2 : add a char c to the end of the string.

Operation 3 : ask the number of different charming substrings.

Operation 4 : ask the number of charming substrings, the same substrings which starts in different location has to be counted.

At the beginning, Victor has an empty string.

 

Input

The input contains several test cases, at most 5 cases.

In each case, the first line has one integer n means the number of operations.

The first number of next n line is the integer op, meaning the type of operation. If op=1 or 2, there will be a lowercase English letters followed.

1≤n≤100000.

 

Output

For each query operation(operation 3 or 4), print the correct answer.

 

Sample Input

6
1 a
1 b
2 a
2 c
3
4
8
1 a
2 a
2 a
1 a
3
1 b
3
4

 

Sample Output

4
5
4
5
11

 

Source

BestCoder Round #52 (div.1) ($)

 

/* ***********************************************
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 suffix

int 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
",p-2);
			else if(kind==4) printf("%lld
",tot);
		}
	}
    
    return 0;
}