【HDU】3518 Boring counting 后缀数组

传送门:【HDU】3518 Boring counting


题目分析:这题呢,我们先构造出后缀数组,求出height数组。然后我们枚举字符串的长度,易知【L,R】内连续的height[i]有共同的前缀,前缀的最长长度为【L+1,R】内所有height的最小值。于是我们可以枚举字符串长度d,然后对该长度找到每一个连续的区间,如果某个区间内最左端l的该前缀和最右端r的该前缀无重叠,即l+d<=r,则++ans。


后缀数组的性质还需要多多掌握啊。


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 = 1005 ;
const int INF = 0x3f3f3f3f ;

char s[MAXN] ;
int t1[MAXN] , t2[MAXN] , c[MAXN] , xy[MAXN] ;
int sa[MAXN] , rank[MAXN] , height[MAXN] ;

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 = 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 , n ) xy[i] = x[y[i]] ;
		rep ( i , 0 , m ) c[i] = 0 ;
		rep ( i , 0 , n ) ++ c[xy[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 solve () {
	int n = strlen ( s ) , ans = 0 ;
	da ( n + 1 ) ;
	For ( i , 1 , n / 2 ) {
		int l = INF , r = -INF ;
		For ( j , 1 , n ) {
			if ( height[j] < i ) {
				if ( r - l >= i ) ++ ans ;
				l = INF , r = -INF ;
			} else {
				l = min ( l , min ( sa[j - 1] , sa[j] ) ) ;
				r = max ( r , max ( sa[j - 1] , sa[j] ) ) ;
			}
		}
		if ( r - l >= i ) ++ ans ;
	}
	printf ( "%d\n" , ans ) ;
}

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


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