【HDU】5343 MZL's Circle Zhou【后缀自动机】

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传送门：【HDU】5343 MZL’s Circle Zhou

对于a串可能和b串重复的部分，我们总能找到一个位置，使得a串达到最长，即a串的后继为空，所以我们只要预处理以字符x为开头的b串的个数即可。

my code:

#include <bits/stdc++.h>using namespace std ;typedef long long LL ;#define clr( a , x ) memset ( a , x , sizeof a )const int MAXN = 100005 ;namespace Suffix_Automation {    const int N = 26 ;    struct Node {        Node* nxt[N] ;        Node* fail ;        LL num ;        int l ;        int c ;        void newnode ( int _l , int _c ) {            clr ( nxt , NULL ) ;            fail = NULL ;            num = 1 ;            l = _l ;            c = _c ;        }    } ;    Node node[MAXN << 1] ;    Node* last ;    Node* root ;    Node* cur ;    void init () {        root = last = cur = node ;        cur->newnode ( 0 , 0 ) ;    }    void create ( Node* p , Node* np , int c ) {        Node *q = p->nxt[c] , *nq = ++ cur ;        *nq = *q ;        nq->l = p->l + 1 ;        q->fail = nq ;        if ( np != NULL ) np->fail = nq ;        for ( ; p && p->nxt[c] == q ; p = p->fail ) p->nxt[c] = nq ;    }    void add ( int c ) {        c -= 'a' ;        Node *p = last , *np = last = ++ cur ;        np->newnode ( p->l + 1 , c ) ;        for ( ; p && p->nxt[c] == NULL ; p = p->fail ) p->nxt[c] = np ;        if ( p == NULL ) np->fail = root ;        else {            if ( p->l + 1 == p->nxt[c]->l ) np->fail = p->nxt[c] ;            else create ( p , np , c ) ;        }    }} ;using namespace Suffix_Automation ;char s1[MAXN] , s2[MAXN] ;int idx[MAXN << 1] ;LL d[26] ;int cmp ( const int& a , const int& b ) {    return node[a].l > node[b].l ;}void solve () {    unsigned long long ans = 1 ;    scanf ( "%s%s" , s1 , s2 ) ;    init () ;    for ( int i = 0 ; s2[i] ; ++ i ) {        add ( s2[i] ) ;    }    int cnt = cur - node ;    for ( int i = 0 ; i <= cnt ; ++ i ) {        idx[i] = i ;    }    sort ( idx , idx + cnt + 1 , cmp ) ;    for ( int i = 0 ; i <= cnt ; ++ i ) {        Node* p = idx[i] + node ;        for ( int j = 0 ; j < N ; ++ j ) {            if ( p->nxt[j] != NULL ) p->num += p->nxt[j]->num ;        }    }    for ( int i = 0 ; i < N ; ++ i ) {        if ( root->nxt[i] != NULL ) d[i] = root->nxt[i]->num ;        else d[i] = 0 ;    }    init () ;    for ( int i = 0 ; s1[i] ; ++ i ) {        add ( s1[i] ) ;    }    cnt = cur - node ;    for ( int i = 0 ; i < N ; ++ i ) {        if ( root->nxt[i] == NULL ) ans += d[i] ;    }    for ( int i = 1 ; i <= cnt ; ++ i ) {        Node* p = node + i ;        unsigned long long l = p->l - p->fail->l ;        for ( int j = 0 ; j < N ; ++ j ) {            if ( p->nxt[j] == NULL ) ans += l * d[j] ;        }        ans += l ;    }    printf ( "%I64u\n" , ans ) ;}int main () {    int T ;    scanf ( "%d" , &T ) ;    for ( int i = 1 ; i <= T ; ++ i ) {        solve () ;    }    return 0 ;}

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