【HDU】4691 Front compression 后缀数组+RMQ

传送门：【HDU】4691 Front compression

题目分析：首先构造好后缀数组，然后对height数组进行rmq预处理，然后每次查询就是O（1）的了。

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 = 100005 ;

char s[MAXN] ;
int t1[MAXN] , t2[MAXN] , c[MAXN] , xy[MAXN] ;
int sa[MAXN] , rank[MAXN] , height[MAXN] ;
int minv[MAXN][17] ;
int n , m ;

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 rmq_init ( int n ) {
	For ( i , 1 , n ) minv[i][0] = height[i] ;
	for ( int j = 1 ; ( 1 << j ) <= n ; ++ j ) {
		for ( int i = 1 ; i + ( 1 << j ) - 1 <= n ; ++ i ) {
			minv[i][j] = min ( minv[i][j - 1] , minv[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 ( minv[L][k] , minv[R - ( 1 << k ) + 1][k] ) ;
}

int lcp ( int a , int b ) {
	if ( a == b ) return n - a ;
	int x = rank[a] , y = rank[b] ;
	return x < y ? rmq ( x + 1 , y ) : rmq ( y + 1 , x ) ;
}

int count ( int n , int n1 = 0 ) {
	for ( n1 = n ? 0 : 1 ; n ; n /= 10 ) ++ n1 ;
	return n1 ;
}

void solve () {
	n = strlen ( s ) ;
	LL ans1 = 0 , ans2 = 0 ;
	int x , y , a , b ;
	da ( n + 1 ) ;
	rmq_init ( n ) ;
	scanf ( "%d" , &m ) ;
	scanf ( "%d%d" , &x , &y ) ;
	ans1 += y - x + 1 ;
	ans2 += y - x + 3 ;//y - x + 2 + count ( 0 ) - 0 ;
	while ( -- m ) {
		scanf ( "%d%d" , &a , &b ) ;
		ans1 += b - a + 1 ;
		int tmp = min ( lcp ( x , a ) , min ( y - x , b - a ) ) ;
		ans2 += b - a + 2 + count ( tmp ) - tmp ;
		x = a , y = b ;
	}
	printf ( "%I64d %I64d\n" , ans1 , ans2 ) ;
}

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