Uvalive 5103 Computer Virus on Planet Pandora Description the number of species that appear in the pattern string AC automata

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Test instructions

Case number

N Mode string

A parent string.

Q: The number of occurrences of a pattern string (a pattern string appears only once)

For "ABC", if the mother string appeared "CBA" Such crossdress, also appeared.

So:

1

Abc

Cba

Ans = 1

#include <cstdio> #include <cstring> #include <cmath> #include <queue> #include <algorithm >using namespace Std;const int maxnode = 250*1000+10000;const int sigma_size = 26;struct Trie{int ch[maxnode][sigma_siz E];int Val[maxnode]; The number of occurrences of the word in the pattern string int last[maxnode];int F[maxnode]; Mismatch array int Num[maxnode];    The number of times the word appears in the text string int pre[maxnode];    The precursor of the word int len[maxnode]; The word length at the end of the word int char[maxnode];   The word corresponds to the letter int Road[maxnode];    Path compression optimization The number of occurrences of the calculation pattern string int sz;int Newnode () {Val[sz] = F[sz] = Last[sz] = Len[sz] = Num[sz] = 0;    memset (Ch[sz], 0, sizeof Ch[sz]);    return sz++;    }void init () {sz=0; Newnode ();} int idx (char c) {return C ' A ';}  int Insert (char *s) {int u = 0;for (int i = 0, C; s[i]; i++) {c = idx (S[i]), if (!ch[u][c]) ch[u][c] = Newnode ();p re[ch[u][c]] = U CHAR[CH[U][C]] = s[i];len[ch[u][c]] = len[u]+1;road[ch[u][c]] = 1;u = Ch[u][c];}   Val[u] = 1;num[u] = 0;return u;}        void Getfail () {queue<int> q; for (int i = 0; i<sigma_size; i++) if (Ch[0][i]) Q.push (Ch[0][i]);        int r, c, u, v;            while (!q.empty ()) {R = Q.front (); Q.pop ();                for (c = 0; c<sigma_size; C + +) {u = ch[r][c];                if (!u) continue;                Q.push (U);                v = f[r]; while (v && ch[v][c] = = 0) v = f[v];            Go up along the mismatch edge if there is a node after mismatch and its child node C exists then end loop f[u] = Ch[v][c]; }}}void find (char *t) {//The number of occurrences of the calculation pattern string: (multiple occurrences per type) int j = 0;for (int i = 0, C, temp; T[i];                 i++) {c = idx (T[i]), while (J && ch[j][c]==0) j = f[j];j = Ch[j][c];temp = J;while (temp) {num[temp]++;            temp = f[temp]; }}}} void Find_kind (char *t, int &ans) {//calculates the number of species, repeated occurrences are no longer computed (if multiple queries are to be added here for (i=0->sz) Lu[i]=1;int j = 0 , I, C, temp;for (i = 0; T[i]; i++) {c = idx (T[i]), while (J && ch[j][c] = = 0) j = f[j];j = Ch[j][c];temp = J;while (temp && road[temp]) {if (V Al[temp]) {++ans;vaL[temp] = 0;}                Road[temp] = 0;            temp = f[temp]; }}}}ac;char s[1015], a[5100010], b[5100010], C;int N, num;int main () {int T; scanf ("%d", &t); while (T--&gt ; 0) {ac.init (); scanf ("%d", &n), while (n--) {scanf ("%s", s); Ac.insert (s);} Ac.getfail (); int top = 0;SCANF ("%s", a), int i = 0;while (A[i]) {if (a[i]!= ' [') b[top++] = a[i];else {num = 0;i++;while (' 0 ' &L t;= A[i] && a[i] <= ' 9 ') num = num*10 + a[i]-' 0 ', i++;c = a[i];i++;while (num--> 0) {b[top++] = c;}} i++;} A[top] = B[top] = 0;for (int i = top-1; I >= 0; i--) a[i] = b[top-i-1];int ans = 0;ac.find_kind (b, ans); Ac.find_kind (A, a NS);p rintf ("%d\n", ans);} return 0;}

Uvalive 5103 Computer Virus on Planet Pandora Description the number of species that appear in the pattern string AC automata