HDOJ-2222 (AC自动机模板)

题目

Problem Description

In the modern time, Search engine came into the life of everybody like Google, Baidu, etc.

Wiskey also wants to bring this feature to his image retrieval system.

Every image have a long description, when users type some keywords to find the image, the system will match the keywords with description of image and show the image which the most keywords be matched.

To simplify the problem, giving you a description of image, and some keywords, you should tell me how many keywords will be match.

Input

First line will contain one integer means how many cases will follow by.

Each case will contain two integers N means the number of keywords and N keywords follow. (N <= 10000)

Each keyword will only contains characters ‘a’-‘z’, and the length will be not longer than 50.

The last line is the description, and the length will be not longer than 1000000.

Output

Print how many keywords are contained in the description.

Sample Input

1

5

she

he

say

shr

her

yasherhs

分析

这就是一个AC自动机的模板题，多组数据。暑假刚学了AC自动机，其实看成是树上的 kmp 就可以了。有时间写个 AC自动机 的笔记:)

先把我的 KMP 的笔记放上来吧，可以看看，当然也可以看其他人的，或许更加清晰。

http://blog.csdn.net/jackypigpig/article/details/54809839

程序

#include <cstdio>
#include <cstring>
using namespace std;
int n,T,ans;
char ch[1000005];

struct AC_automaton{
#define X (x[i]-'a')
#define son(x) to[fa][x]
struct node{int boo;} t[1000005];
int i,j,L,k,fa,cnt,to[1000005][26],fail[1000005],l,r,q[1000005];

void init(){
memset(t,0,sizeof(t));
memset(to,0,sizeof(to));
memset(fail,0,sizeof(fail));
cnt=ans=0;
}

void insert(char *x){
L=strlen(x);
for (fa=i=0; i<L; i++) fa=(son(X)?son(X):(son(X)=++cnt));
t[fa].boo++;
}

void getfail(){
for (i=q[l=r=0]; l<=r; i=q[++l])
for (j=0; j<26; j++)
if (to[i][j]){
q[++r]=to[i][j];
for (fa=fail[i]; fa && !son(j); fa=fail[fa]);
if (i>0 && son(j)) fail[to[i][j]]=son(j);
}
else to[i][j]=to[fail[i]][j];
}

void query(char *x){
L=strlen(x);
for (fa=i=0; i<L; i++)
for (k=fa=son(X); k && t[k].boo!=-1; k=fail[k]){
ans+=t[k].boo;
t[k].boo=-1;
}
}

} AC;

int main(){
freopen("1.txt","r",stdin);
for (scanf("%d",&T); T--;){
scanf("%d",&n);
for (int i=1; i<=n; i++) scanf("%s",ch),AC.insert(ch);
AC.getfail();
scanf("%s",ch);
AC.query(ch);
printf("%d\n",ans);
AC.init();
}
}

提示

同样的程序（去掉多组数据的那几行）在 luogu-3808 上也能交。