Keywords Search AC自动机
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.
InputFirst 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.
OutputPrint how many keywords are contained in the description.Sample Input
1
5
she
he
say
shr
her
yasherhs
Sample Output
3
AC自动机:利用ttie树节省空间,利用kmp相似的原理构建fail指针进行匹配
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<sstream>
#include<algorithm>
#include<queue>
#include<deque>
#include<iomanip>
#include<vector>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<fstream>
#include<memory>
#include<list>
#include<string>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define MAXN 26
#define L 31
#define INF 1000000009
#define eps 0.00000001
struct node
{
node()
{
cnt = ;
fail = NULL;
for (int i = ; i < MAXN; i++)
next[i] = NULL;
}
int cnt;
struct node* next[MAXN];
struct node* fail;
};
typedef struct node* Tree;
void insert(Tree T, const char* str)
{
int p = ;
Tree tmp = T;
while (str[p] != '\0')
{
int k = str[p] - 'a';
if (tmp->next[k] == NULL)
{
tmp->next[k] = new node();
}
tmp = tmp->next[k];
p++;
}
tmp->cnt++;
}
void build_ac(Tree root)
{
root->fail = NULL;
queue<Tree> q;
q.push(root);
while (!q.empty())
{
Tree tmp = q.front();
q.pop();
Tree p = NULL;
for (int i = ; i < MAXN; i++)
{
if (tmp->next[i] != NULL)
{
if (tmp == root)
tmp->next[i]->fail = root;
else
{
p = tmp->fail;
while (p != NULL)
{
if (p->next[i] != NULL)
{
tmp->next[i]->fail = p->next[i];
break;
}
p = p->fail;
}
if (p == NULL) tmp->next[i]->fail = root;
}
q.push(tmp->next[i]);
}
}
}
}
int query(Tree root,const char* str)
{
int i = ;
int count = ;
int l = strlen(str);
Tree tmp = root;
int index;
while (str[i])
{
index = str[i] - 'a';
while (tmp->next[index] == NULL && tmp != root)
tmp = tmp->fail;
tmp = tmp->next[index];
if (tmp == NULL) tmp = root;
Tree p = tmp;
while (p != root)
{
count += p->cnt;
p->cnt = ;
p = p->fail;
}
i++;
}
return count;
} void del(Tree root)
{
for (int i = ; i < MAXN; i++)
if (root->next[i] != NULL)
del(root->next[i]);
delete root;
}
char key[], s[];
int main()
{
int T, n;
scanf("%d", &T);
while (T--)
{
scanf("%d", &n);
Tree root = new node();
root->cnt = ;
root->fail = NULL;
for (int i = ; i < MAXN; i++)
root->next[i] = NULL;
for (int i = ; i < n; i++)
{
scanf("%s", key);
insert(root, key);
}
build_ac(root);
scanf("%s", s);
printf("%d\n", query(root, s));
del(root);
}
}
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