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【ural】1960. Palindromes and Super Abilities【Palindromic Tree】

2014-12-23 17:14 435 查看
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题目分析:这题就是裸回文树了,按照读入顺序边插入边输出答案即可。。答案就是插入当前字符后回文树内表示回文串的节点个数。

代码如下:

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;

typedef long long LL ;

#define rep( i , a , b ) for ( int i = ( a ) ; i <  ( b ) ; ++ i )
#define For( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )
#define rev( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i )
#define clr( a , x ) memset ( a , x , sizeof a )

const int MAXN = 100005 ;
const int N = 26 ;

struct Palindromic_Tree {
	int next[MAXN]
 ;
	int fail[MAXN] ;
	int cnt[MAXN] ;
	int len[MAXN] ;
	int S[MAXN] , n ;
	int last ;
	int p ;

	int newnode ( int l ) {
		rep ( i , 0 , N ) next[p][i] = 0 ;
		cnt[p] = 0 ;
		len[p] = l ;
		return p ++ ;
	}

	void init () {
		p = 0 ;
		newnode (  0 ) ;
		newnode ( -1 ) ;
		last = 0 ;
		n = 0 ;
		S
 = -1 ;
		fail[0] = 1 ;
	}

	int get_fail ( int x ) {
		while ( S[n - len[x] - 1] != S
 ) x = fail[x] ;
		return x ;
	}

	int add ( int c ) {
		int ret = 0 ;
		c -= 'a' ;
		S[++ n] = c ;
		int cur = get_fail ( last ) ;
		if ( !next[cur][c] ) {
			ret = 1 ;
			int now = newnode ( len[cur] + 2 ) ;
			fail[now] = next[get_fail ( fail[cur] )][c] ;
			next[cur][c] = now ;
		}
		last = next[cur][c] ;
		cnt[last] ++ ;
		return ret ;
	}

	void count () {
		for ( int i = p - 1 ; i >= 0 ; -- i ) cnt[fail[i]] += cnt[i] ;
	}
} ;

Palindromic_Tree T ;
char s[MAXN] ;
int n ;

void solve () {
	int ans = 0 ;
	T.init () ;
	n = strlen ( s ) ;
	rep ( i , 0 , n ) printf ( "%d%c" , ans += T.add ( s[i] ) , i < n - 1 ? ' ' : '\n' ) ;
}

int main () {
	while ( ~scanf ( "%s" , s ) ) solve () ;
	return 0 ;
}
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