【POJ】1743 Musical Theme 【后缀数组】

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传送门：【POJ】1743 Musical Theme

题目分析：这题和HDU3518做法差不多。首先将序列差分，然后构造后缀数组，接下来我们二分重复的串的长度k，然后看height[i]连续大于等于k的里面的最左端L是否和最右端重叠，如果不重叠则修改下界，否则如果找不到则修改上界。最后看找到的长度是否大于等于4（差分后少了一个数），有的话输出这个长度，否则输出0。

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 = 20005 ;int s[MAXN] ;int t1[MAXN] , t2[MAXN] , c[MAXN] , xy[MAXN] ;int sa[MAXN] , rank[MAXN] , height[MAXN] ;int n ;bool 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 = 200 ) {int i , d = 1 , p = 0 , *x = t1 , *y = t2 , *t ;for ( i = 0 ; i < m ; ++ i ) c[i] = 0 ;for ( i = 0 ; i < n ; ++ i ) ++ c[x[i] = s[i]] ;for ( i = 1 ; i < m ; ++ i ) c[i] += c[i - 1] ;for ( i = n - 1 ; i >= 0 ; -- i ) sa[-- c[x[i]]] = i ;for ( ; p < n ; d <<= 1 , m = p ) {for ( p = 0 , i = n - d ; i < n ; ++ i ) y[p ++] = i ;for ( i = 0 ; i < n ; ++ i ) if ( sa[i] >= d ) y[p ++] = sa[i] - d ;for ( i = 0 ; i < m ; ++ i ) c[i] = 0 ;for ( i = 0 ; i < n ; ++ i ) ++ c[xy[i] = x[y[i]]] ;for ( i = 1 ; i < m ; ++ i ) c[i] += c[i - 1] ;for ( i = n - 1 ; i >= 0 ; -- i ) sa[-- c[xy[i]]] = y[i] ;for ( t = x , x = y , y = t , p = 1 , x[sa[0]] = 0 , i = 1 ; i < n ; ++ i ) {x[sa[i]] = cmp ( y , sa[i - 1] , sa[i] , d ) ? p - 1 : p ++ ;}}getHeight ( n - 1 ) ;}int check ( int k ) {int L = MAXN , R = -1 ;For ( i , 2 , n ) {if ( height[i] < k ) {if ( L + k <= R ) return 1 ;L = MAXN , R = -1 ;} else {L = min ( L , min ( sa[i - 1] , sa[i] ) ) ;R = max ( R , max ( sa[i - 1] , sa[i] ) ) ;}}return L + k <= R ;}void solve () {rep ( i , 0 , n ) scanf ( "%d" , &s[i] ) ;-- n ;rep ( i , 0 , n ) s[i] = s[i + 1] - s[i] + 100 ;da ( n + 1 ) ;int l = 0 , r = n ;while ( l < r ) {int m = ( l + r + 1 ) >> 1 ;if ( check ( m ) ) l = m ;else r = m - 1 ;}if ( l >= 4 ) printf ( "%d\n" , l + 1 ) ;else printf ( "0\n" ) ;}int main () {while ( ~scanf ( "%d" , &n ) && n ) solve () ;return 0 ;}

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