【URAL】1297 Palindrome 【后缀数组+RMQ——求最长回文子串】

传送门：【URAL】1297 Palindrome

题目分析：将s串倒过来接到原串的后面，中间用'$'隔开，然后我们构造后缀数组和height数组，接着我们RMQ预处理，接下来枚举串的每个点作为回文串的中心，分别以该点为奇回文串中心以及偶回文串中心求LCP，此时向左和向右的串的LCP即以这个点作为中心的回文串的扩展半径，求半径的话我们就对原串位置以及倒过来后的相应位置做LCP。

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 )#define cpy( a , x ) memcpy ( a , x , sizeof a )const int MAXN = 2005 ;char s[MAXN] ;int t1[MAXN] , t2[MAXN] , c[MAXN] , xy[MAXN] ;int sa[MAXN] , rank[MAXN] , height[MAXN] ;int dp[MAXN][12] ;int cmp ( int *r , int a , int b , int d ) {return r[a] == r[b] && r[a + d] == r[b + d] ;}void getHeight ( int n , int k = 0 ) {For ( i , 0 , n ) rank[sa[i]] = i ;rep ( i , 0 , n ) {if ( k ) -- k ;int j = sa[rank[i] - 1] ;while ( s[i + k] == s[j + k] ) ++ k ;height[rank[i]] = k ;}}void da ( int n , int m = 128 ) {int *x = t1 , *y = t2 ;rep ( i , 0 , m ) c[i] = 0 ;rep ( i , 0 , n ) ++ c[x[i] = s[i]] ;rep ( i , 1 , m ) c[i] += c[i - 1] ;rev ( i , n - 1 , 0 ) sa[-- c[x[i]]] = i ;for ( int d = 1 , p = 0 ; p < n ; d <<= 1 , m = p ) {p = 0 ;rep ( i , n - d , n ) y[p ++] = i ;rep ( i , 0 , n ) if ( sa[i] >= d ) y[p ++] = sa[i] - d ;rep ( i , 0 , m ) c[i] = 0 ;rep ( i , 0 , n ) ++ c[xy[i] = x[y[i]]] ;rep ( i , 1 , m ) c[i] += c[i - 1] ;rev ( i , n - 1 , 0 ) sa[-- c[xy[i]]] = y[i] ;swap ( x , y ) ;p = 0 ;x[sa[0]] = p ++ ;rep ( i , 1 , n ) x[sa[i]] = cmp ( y , sa[i - 1] , sa[i] , d ) ? p - 1 : p ++ ;}getHeight ( n - 1 ) ;}void init_RMQ ( int n ) {For ( i , 1 , n ) dp[i][0] = height[i] ;for ( int j = 1 ; ( 1 << j ) <= n ; ++ j ) {for ( int i = 1 ; i + ( 1 << j ) - 1 <= n ; ++ i ) {dp[i][j] = min ( dp[i][j - 1] , dp[i + ( 1 << ( j - 1 ) )][j - 1] ) ;}}}int RMQ ( int L , int R ) {int k = 0 ;while ( ( 1 << ( k + 1 ) ) <= R - L + 1 ) ++ k ;return min ( dp[L][k] , dp[R - ( 1 << k ) + 1][k] ) ;}int lcp ( int a , int b ) {a = rank[a] , b = rank[b] ;return a < b ? RMQ ( a + 1 , b ) : RMQ ( b + 1 , a ) ;}void solve () {int n1 = strlen ( s ) ;int n2 = n1 ;int n = n1 + 1 + n2 ;s[n1] = '$' ;s[n] = 0 ;rep ( i , 0 , n1 ) s[n - i - 1] = s[i] ;da ( n + 1 ) ;init_RMQ ( n ) ;int t = 0 , ans = 0 ;rep ( i , 0 , n1 ) {int x = lcp ( i , n - i - 1 ) ;if ( x * 2 - 1 > t ) {t = x * 2 - 1 ;ans = i - x + 1 ;}int y = lcp ( i , n - i ) ;if ( y * 2 > t ) {t = y * 2 ;ans = i - y ;}}rep ( i , ans , ans + t ) printf ( "%c" , s[i] ) ;printf ( "\n" ) ;}int main () {while ( ~scanf ( "%s" , s ) ) solve () ;return 0 ;}