【HDU】4552 怪盗基德的挑战书 【后缀数组】

传送门：【HDU】4552 怪盗基德的挑战书

题目分析：答案就是所有后缀和串s的lcp长度相加。后缀数组搞定。

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 dp[MAXN][17] ;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 x , int y ) {int k = 0 ;while ( ( 1 << ( k + 1 ) ) <= y - x + 1 ) ++ k ;return min ( dp[x][k] , dp[y - ( 1 << k ) + 1][k] ) ;}int lcp ( int x , int y ) {x = rank[x] ;y = rank[y] ;return x < y ? rmq ( x + 1 , y ) : rmq ( y + 1 , x ) ;}void solve () {int n = strlen ( s ) ;int ans = n , minv = MAXN ;da ( n + 1 ) ;init_RMQ ( n ) ;rep ( i , 1 , n ) ans = ( ans + lcp ( 0 , i ) ) % 256 ;printf ( "%d\n" , ans ) ;}int main () {while ( ~scanf ( "%s" , s ) ) solve () ;return 0 ;}

上面的代码用了RMQ，但实际上是不需要的，那是我对后缀数组的不了解造成的。其实只要从rank[0]向两边扫就OK了的。

代码如下：

void solve () {int n = strlen ( s ) ;int ans = n ;da ( n + 1 ) ;int x = rank[0] , y = x , t = MAXN ;while ( y > 0 ) {t = min ( t , height[y --] ) ;ans = ( ans + t ) % 256 ;}y = x + 1 , t = MAXN ;while ( y <= n ) {t = min ( t , height[y ++] ) ;ans = ( ans + t ) % 256 ;}printf ( "%d\n" , ans ) ;}