POJ1509 Glass Beads (后缀自动机)

题意：一个字符串可以将第一个字符放到最后一位，然后问不断这样做可以得到的字典序最小的字符串

#include <stdio.h>
#include <string.h>
#include <set>
#include <algorithm>
using namespace std ;

typedef long long LL ;
typedef unsigned long long ULL ;
typedef pair < int , int > pii ;

#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 )

const int MAXN = 40005 ;
const int W = 26 ;

struct Node {
    Node* next[W] ;
    Node* fail ;
    int len ;
    int l ;
    void newnode () {
        clr ( next , 0 ) ;
        fail = NULL ;
        len = l = 0 ;
    }
} ;

typedef Node* P_Node ;

struct SAM {
    Node node[MAXN] ;
    P_Node last , root ;
    int cnt ;

    void init () {
        cnt = 0 ;
        node->newnode () ;
        root = last = node ;
    }

    void add ( int c ) {
        P_Node p = last , np = node + ( ++ cnt ) ;
        np->newnode () ;
        np->l = np->len = p->len + 1 ;
        last = np ;
        for ( ; p && !p->next[c] ; p = p->fail ) p->next[c] = np ;
        if ( !p ) np->fail = root ;
        else {
            if ( p->len + 1 == p->next[c]->len ) np->fail = p->next[c] ;
            else {
                P_Node q = p->next[c] , nq = node + ( ++ cnt ) ;
                *nq = *q ;
                nq->len = p->len + 1 ;
                nq->l = np->len ;
                q->fail = np->fail = nq ;
                for ( ; p && p->next[c] == q ; p = p->fail ) p->next[c] = nq ;
            }
        }
    }
} ;

SAM sam ;
char s[MAXN] ;
int n ;

void solve () {
    sam.init () ;
    scanf ( "%s" , s ) ;
    n = strlen ( s ) ;
    rep ( i , 0 , n ) sam.add ( s[i] - 'a' ) ;
    rep ( i , 0 , n ) sam.add ( s[i] - 'a' ) ;
    P_Node now = sam.root ;
    rep ( i , 0 , n ) {
        rep ( j , 0 , W ) if ( now->next[j] ) {
            now = now->next[j] ;
            break ;
        }
    }
    printf ( "%d
" , now->l - n + 1 ) ;
}

int main () {
    int T ;
    scanf ( "%d" , &T ) ;
    For ( i , 1 , T ) solve () ;
    return 0 ;
}