hdu 2222 ac自动机入门题 可以做模板

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 25141    Accepted Submission(s): 8224


[align=left]Problem Description[/align]
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.

 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
Print how many keywords are contained in the description.
 

[align=left]Sample Input[/align]

1
5
she
he
say
shr
her
yasherhs

 

[align=left]Sample Output[/align]

3

 

[align=left]Author[/align]
Wiskey
 

[align=left]Recommend[/align]
lcy
 
 
http://acm.hdu.edu.cn/showproblem.php?pid=2222
 
 
建议不了解ac自动机的 请看看
http://hi.baidu.com/nialv7/item/ce1ce015d44a6ba7feded52d
http://www.cppblog.com/mythit/archive/2009/04/21/80633.html
 
题意：

题意：给出n个串，然后给一篇文章，问这n个串有多少个在文章里面出现过。。。

trick：n个串可能有相同的，需按照不同串处理。

 

#include<stdio.h>
#include<string.h>
#include<malloc.h>
#include<queue>
using namespace std;
char str[1000000+100];

struct node
{
int count;
struct node *next[26];
struct node *fail;
void init()
{
int i;
for(i=0;i<26;i++)
next[i]=NULL;
count=0;
fail=NULL;
}
}*root;
void insert()
{
int len,k;
node *p=root;
len=strlen(str);
for(k=0;k<len;k++)
{
int pos=str[k]-'a';
if(p->next[pos]==NULL)
{
p->next[pos]=new node;
p->next[pos]->init();
p=p->next[pos];
}
else
p=p->next[pos];
}
p->count++;
}
void getfail()
{
int i;
node *p=root,*son,*temp;
queue<struct node *>que;
que.push(p);
while(!que.empty())
{
temp=que.front();
que.pop();
for(i=0;i<26;i++)
{
son=temp->next[i];
if(son!=NULL)
{
if(temp==root) {son->fail=root;}
else
{
p=temp->fail;
while(p)
{
if(p->next[i])
{
son->fail=p->next[i];
break;
}
p=p->fail;
}
if(!p)  son->fail=root;
}
que.push(son);
}
}
}
}
void query()
{
int len,i,cnt=0;
len=strlen(str);
node *p,*temp;
p=root;
for(i=0;i<len;i++)
{
int pos=str[i]-'a';
while(!p->next[pos]&&p!=root)  p=p->fail;
p=p->next[pos];//
if(!p) p=root;//
temp=p;
/*不要用*temp=*p  因为*p表示一个node，而*temp也表示一个node 但是由于*temp没有分配空间 所以是不能进行赋值的 但是可以用temp指针去指向p*/
while(temp!=root)
{
if(temp->count>=0)
{
cnt+=temp->count;
temp->count=-1;
}
else break;
temp=temp->fail;
}
}
printf("%d\n",cnt);
}
int main()
{
int cas,n;
scanf("%d",&cas);
while(cas--)
{
root=new node;
root->init();
root->fail=NULL;
scanf("%d",&n);
int i;
getchar();
for(i=0;i<n;i++)
{
gets(str);
insert();
}
getfail();
gets(str);
query();
}
return 0;
}