Rokua P3808 Template AC automaton (simple version)

Topic background

This is a simple AC automaton template problem.

Used to detect correctness and algorithm constants.

In order to prevent card OJ, only two sets of data are guaranteed on the correct basis, please do not commit maliciously.

Title Description

Given n pattern strings and a text string, how many pattern strings appear in the text string.

Input/output format

Input format:
The first row is an n, which indicates the number of pattern strings;

The following n rows a pattern string per line;

The following line is a text string.

Output format:
A number means an answer.

Input/Output sample

Input Sample # #:
2
A
Aa
Aa
Sample # # of output:
2
Description

Subtask1[50pts]:∑length (Pattern String) <=10^6,length (text string) <=10^6,n=1;

Subtask2[50pts]:∑length (Pattern String) <=10^6,length (text string) <=10^6;

Exercises
AC Automaton Bare Topic

Code (trie Chart)

#include <cstdio> #include <cstring> #include <iostream> using namespace std;
    inline int read () {int X=0,f=1;char ch=getchar (); while (ch< ' 0 ' | |
    Ch> ' 9 ') {if (ch== '-') F=-1;ch=getchar ();}
    while (ch>= ' 0 ' &&ch<= ' 9 ') {x=x*10+ch-' 0 '; Ch=getchar ();}
return x*f;
} Char s[1000001];
int T,n,tot,ans;
int a[1000001][26],q[1000001],point[1000001],danger[1000001];
    inline void ins () {int Now=1,l=strlen (s);
        for (int i=0;i<l;i++) {int t=s[i]-' a ';
        if (!a[now][t]) A[now][t]=++tot;
    NOW=A[NOW][T];
} danger[now]++;
    } inline void Acmach () {int head=0,tail=1;
    q[1]=1;point[1]=0;
        while (head!=tail) {int now=q[++head]; for (int i=0;i<26;i++) {if (A[now][i]) {Point[a[now][i]]=a[point[now]
                ][i];
            Q[++tail]=a[now][i];
        } else a[now][i]=a[point[now]][i]; }}} inline void Solve () {inT Now=1,l=strlen (s);
        for (int i=0;i<l;i++) {int t=s[i]-' a ';
        NOW=A[NOW][T];
    for (int j=now;j&&danger[j]!=-1;j=point[j]) ans+=danger[j],danger[j]=-1;
} printf ("%d", ans);
    } int main () {for (int i=0;i<26;i++) a[0][i]=1;
    N=read (); tot=1;
        while (n--) {scanf ("%s", s);
    Ins ();
    } Acmach ();
    scanf ("%s", s);
    Solve ();
return 0; }