bzoj3473: 字符串

Description
给定n个字符串，询问每个字符串有多少子串（不包括空串）是所有n个字符串中至少k个字符串的子串？
Input
第一行两个整数n，k。
接下来n行每行一个字符串。
Output
一行n个整数，第i个整数表示第i个字符串的答案。
Sample Input
3 1
abc
a
ab
Sample Output
6 1 3
HINT
对于 100% 的数据，1<=n，k<=10^5，所有字符串总长不超过10^5，字符串只包含小写字母。
Source
Adera 1 杯冬令营模拟赛

用广义SAM做..

用set维护到达当前点所属串的个数，顺着fail启发式合并到根

询问的时候，每走到一个点表示的是这个串的前缀，而我们要统计的就是这个前缀最长后缀使得这个后缀符合条件，然后ans加上这个最长后缀长度就好了，因为1~最长后缀都可以到达..

#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#include <set>#define LL long longusing namespace std;const LL Maxn = 100010;const LL lg = 20;LL F[Maxn*2], d[Maxn*2], ch[Maxn*2][26], tot, now, f[Maxn*2];char s[Maxn*2]; LL len;LL Rsort[Maxn*2], rk[Maxn*2];LL n, K;set <LL> S[Maxn*2];set <LL> :: iterator it;LL copy ( LL p, LL c ){    LL x = ++tot, y = ch[p][c];    d[x] = d[p]+1;    for ( LL i = 0; i < 26; i ++ ) ch[x][i] = ch[y][i];    F[x] = F[y]; F[y] = x;    while ( ~p && ch[p][c] == y ){ ch[p][c] = x; p = F[p]; }    return x;}void add ( LL c ){    LL p, o;    if ( p = ch[now][c] ){        if ( d[p] != d[now]+1 ) copy ( now, c );        now = ch[now][c];    }    else {        d[o=++tot] = d[now]+1; p = now; now = o;        while ( ~p && !ch[p][c] ){ ch[p][c] = o; p = F[p]; }        F[o] = ~p ? ( d[p]+1 == d[ch[p][c]] ? ch[p][c] : copy ( p, c ) ) : 0;    }}void merge ( LL x, LL y ){    while ( S[y].begin () != S[y].end () ){        it = S[y].begin ();        S[x].insert (*it);        S[y].erase (it);    }}LL _max ( LL x, LL y ){ return x > y ? x : y; }int main (){    LL i, j, k;    scanf ( "%lld%lld", &n, &K );    len = 0;    for ( i = 1; i <= n; i ++ ){        scanf ( "%s", s+len );        len = strlen (s);        s[len++] = '$';    }    F[0] = -1; now = 0;    LL ss = 0;    for ( i = 0; i < len; i ++ ){        if ( s[i] != '$' ){ add (s[i]-'a'); S[now].insert (ss); }        else { now = 0; ss ++; }    }    for ( i = 1; i <= tot; i ++ ) Rsort[d[i]] ++;    for ( i = 1; i <= len; i ++ ) Rsort[i] += Rsort[i-1];    for ( i = tot; i >= 1; i -- ) rk[Rsort[d[i]]--] = i;    for ( i = tot; i >= 1; i -- ){        if ( S[rk[i]].size () >= K ) f[rk[i]] = d[rk[i]];        merge ( F[rk[i]], rk[i] );    }    for ( i = 1; i <= tot; i ++ ) f[rk[i]] = _max ( f[F[rk[i]]], f[rk[i]] );    j = 0;    for ( i = 1; i <= n; i ++ ){        if ( s[j] == '$' ) j ++;        now = 0; LL ans = 0;        while ( s[j] != '$' ){            LL c = s[j]-'a';            now = ch[now][c];            ans += f[now];            j ++;        }        printf ( "%lld ", ans );    }    printf ( "\n" );    return 0;}