hdu 3695 AC自动机模板题

Keywords Search

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

Total Submission(s): 23502 Accepted Submission(s): 7763



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

Sample Output

3

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模板题，不能更水。

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#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cstring>

using namespace std;

//子树节点是在插入时new的，
//寻找失配指针中使用的队列是直接调用STL的
const int kind = 26;
struct node
{
node *fail;
node *next[kind];
int count;//记录当前前缀是完整单词出现的个数
node()
{
fail = NULL;
count = 0;
memset(next,NULL,sizeof(next));
}
};

void insert(char *str,node *root)
{
node *p=root;
int i=0,index;
while(str[i])
{
index = str[i]-'a';
if(p->next[index]==NULL) p->next[index]=new node();
p=p->next[index];
i++;
}
p->count++;

}

//寻找失败指针
void build_ac_automation(node *root)
{
int i;
queue<node *>Q;
root->fail = NULL;
Q.push(root);
while(!Q.empty())
{
node *temp = Q.front();//q[head++];//取队首元素
Q.pop();
node *p = NULL;
for(i=0; i<kind; i++)
{
if(temp->next[i]!=NULL)//寻找当前子树的失败指针
{
p = temp->fail;
while(p!=NULL)
{
if(p->next[i]!=NULL)//找到失败指针
{
temp->next[i]->fail = p->next[i];
break;
}
p = p->fail;
}

if(p==NULL)//无法获取，当前子树的失败指针为根
temp->next[i]->fail = root;

Q.push(temp->next[i]);
}
}
}
}

//询问str中包含n个关键字中多少种即匹配
int query(char *str,node *root)
{
int i = 0,cnt = 0,index,len;
len = strlen(str);
node *p = root;
while(str[i])
{
index = str[i]-'a';
while(p->next[index]==NULL&&p!=root)//失配
p=p->fail;
p=p->next[index];
if(p==NULL)//失配指针为根
p = root;

node *temp = p;
while(temp!=root&&temp->count!=-1)//寻找到当前位置为止是否出现病毒关键字
{
//if(temp->count!=0)
cnt+=temp->count;
temp->count=-1;
temp=temp->fail;
}
i++;
}
return cnt;
}

char str[1111111];
char words[1111111];
int T,n;

node* root;

int main()
{
scanf("%d",&T);
while (T--)
{
scanf("%d",&n);
root=new node();
while (n--)
{
scanf("%s",words);
insert(words,root);
}
build_ac_automation(root);
scanf("%s",str);
int ans=query(str,root);
printf("%d\n",ans);
}
return 0;
}