"Bzoj 3879" SvT

http://www.lydsy.com/JudgeOnline/problem.php?id=3879

The Chinese of SVT is suffix virtual tree?

Anyway this konjac konjac do not understand, or $o (NLOGN) $ suffix array and the monotonic stack maintenance to do, Fye learned elder sister said this study method (at that time did not understand Qwq), Xiaoyimi taught me this practice →xiaoyimi the solution.

First contributed 2 times tle, thought it was the death circle of metaphysics, decisively hung up on the camera for a night, and then on the way home to think of Tle is because the time of submission forgot to delete freopen

#include <cstdio> #include <cstring> #include <algorithm>using namespace Std;typedef long ll;    const int N = 500003;const ll p = 23333333333333333ll;int in () {int k = 0, fh = 1; char c = GetChar ();    for (; c < ' 0 ' | | c > ' 9 '; c = GetChar ()) if (c = = '-') FH =-1;    for (; c >= ' 0 ' && C <= ' 9 '; c = GetChar ()) K = (k << 3) + (k << 1) + C-' 0 ';  return k * FH;  int c[n], t1[n], t2[n];    void St (int *x, int *y, int *sa, int n, int m) {for (int i = 0; i < m; ++i) c[i] = 0;    for (int i = 0; i < n; ++i) ++c[x[y[i]];    for (int i = 1; i < m; ++i) c[i] + = c[i-1];  for (int i = n-1; I >= 0; i) sa[--c[x[y[i]]] [= Y[i];}    void mkhz (int *r, int *sa, int n, int m) {int *x = t1, *y = t2, *t, p, I, J;    for (i = 0; i < n; ++i) x[i] = R[i], y[i] = i;    St (x, y, SA, N, M);        for (P = 1, j = 1; p < n; m = p, j <<= 1) {for (P = 0, i = n-j; i < n; ++i) y[p++] = i; for (i = 0; I &Lt N        ++i) if (Sa[i] >= j) y[p++] = sa[i]-J;        St (x, y, SA, N, M); for (t = x, x = y, y = t, p = 1, x[sa[0]] = 0, i = 1; i < n; ++i) x[sa[i]] = y[sa[i] [= y[sa[i-1]] &AMP;&A mp Y[sa[i] + j] = = Y[sa[i-1] + j]?    P-1: p++;    }} void Mkh (int *r, int *sa, int *rank, int *h, int n) {int k = 0, J, I;    for (i = 0; i < n; ++i) rank[sa[i]] = i;  for (i = 1; l < n; h[rank[i++]] = k) for (K---k:0, j = sa[rank[i]-1]; R[i + K] = = R[j + K]; ++k);}  Char S[n];int N, M, Sa[n], rank[n], r[n], h[n], f[n][20], log_2[n], a[n];    int get_min (int l, int r) {int len = log_2[r-l];  return min (F[l][len], F[r-(1 << len)][len]);}    int GET_LCP (int l, int r) {L = Rank[l]; r = Rank[r];    if (L > R) Swap (L, R);  Return Get_min (L, r);}  BOOL CMP (int x, int y) {return rank[x] < rank[y];}  int Sta[n], top, bef[n], size[n];  void Sub (ll &x, ll y) {x-= y; if (x < 0) x + = p;} void Add (ll &x, ll y) {x + = y;if (x > P) x-= p;}    int main () {n = in (); m = in ();    scanf ("%s", S + 1);    R[0] = 0;    for (int i = 1; I <= n; ++i) r[i] = s[i]-' a ' + 1;    Mkhz (R, SA, n + 1, 27);          Mkh (R, SA, rank, h, n + 1);    for (int i = 0; i < n; ++i) f[i][0] = h[i + 1]; for (int j = 1, J < ++j) for (int i = 0; i < n; ++i) {if (i + (1 << (j-1)) >= N) Bre            Ak        F[i][j] = min (f[i][j-1], f[i + (1 << (j-1))][j-1]);    } int tmp = 0;        for (int i = 1; I <= n; ++i) {if ((1 << (tmp + 1)) < i) ++tmp;    Log_2[i] = tmp; } int tot;    ll sum, ans = 0;        while (m--) {tot = in ();        for (int i = 1; I <= tot; ++i) a[i] = in ();        Sort (A + 1, a + tot + 1, CMP);        tot = unique (A + 1, a + tot + 1)-A;        --tot;                  for (int i = 1; i < tot; ++i) bef[i] = GET_LCP (A[i], a[i + 1]); top = 0; sum = 0;        Ans = 0;    for (int i = 1; i < tot; ++i) {        Size[i] = 1;                while (top && bef[i] < Bef[sta[top]) {sub (sum, 1LL * bef[sta[top] [size[sta[top]]% p);                Size[i] + = Size[sta[top]];            --top;            } Sta[++top] = i;            Add (Sum, 1LL * bef[i] * size[i]% P);        Add (ans, sum);    } printf ("%lld\n", ans); } return 0;}

Once again, I was moved by my IQ Qaq

"Bzoj 3879" SvT