Victor and String

题目描述

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.

输入描述:

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.       $1\leq n \leq 100000$.

输出描述:

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

示例1

输入

61 a1 b2 a2 c3481 a2 a2 a1 a31 b34

输出

4545
11
#include <stdio.h>#include <string.h>#include <algorithm>using namespace std ; typedef long long LL ; #define clr( a , x ) memset ( a , x , sizeof a ) const int MAXN = 200005 ; struct Palindromic_Tree {    int nxt[MAXN][26] ;    int fail[MAXN] ;    int len[MAXN] ;    int S[MAXN] , L , R ;    int num[MAXN] ;    int last[2] ;    int p ;    LL tot ;     int newnode ( int l ) {        for ( int i = 0 ; i < 26 ; ++ i ) nxt[p][i] = 0 ;        num[p] = 0 ;        len[p] = l ;        return p ++ ;    }     void init ( int n ) {        p = 0 ;        newnode ( 0 ) ;        newnode ( -1 ) ;         clr ( S , -1 ) ;        L = n ;        R = n - 1 ;        last[0] = last[1] = 1 ;        fail[0] = 1 ;         tot = 0 ;    }     int get_fail ( int v , bool d ) {        if ( d ) while ( S[R - len[v] - 1] != S[R] ) v = fail[v] ;        else while ( S[L + len[v] + 1] != S[L] ) v = fail[v] ;        return v ;    }     void add ( int c , bool d ) {        if ( d ) S[++ R] = c ;        else S[-- L] = c ;        int cur = get_fail ( last[d] , d ) ;        if ( !nxt[cur][c] ) {            int now = newnode ( len[cur] + 2 ) ;            fail[now] = nxt[get_fail ( fail[cur] , d )][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]] ;    }} ; Palindromic_Tree T ;int n ; void solve () {    int op ;    char c ;    T.init ( n ) ;    for ( int i = 0 ; i < n ; ++ i ) {        scanf ( "%d" , &op ) ;        if ( op <= 2 ) {            scanf ( " %c" , &c ) ;            T.add ( c - 'a' , op - 1 ) ;        }        if ( op == 3 ) printf ( "%d\n" , T.p - 2 ) ;        if ( op == 4 ) printf ( "%I64d\n" , T.tot ) ;    }} int main () {    while ( ~scanf ( "%d" , &n ) ) solve () ;    return 0 ;}