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 ; }