hdu:2222:Keywords Search（AC自动机模板题）

Posted 2020-10-03 HWIM

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 66208    Accepted Submission(s): 22193

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

 

 

code

  1 #include<cstdio>
  2 #include<algorithm>
  3 #include<cstring>
  4 #include<queue>
  5 
  6 using namespace std;
  7 
  8 const int MAXN = 500100;
  9 
 10 char str[1000100],s[110];
 11 
 12 struct Aczdj{
 13     int ch[MAXN][26],val[MAXN],last[MAXN],fail[MAXN],size,ret;
 14     void clear()
 15     { 
 16         memset(ch[0],0,sizeof(ch[0]));
 17         memset(val,0,sizeof(val));
 18         memset(fail,0,sizeof(fail));
 19         size = 0;
 20         ret = 0;
 21     }
 22     int idx(char c)
 23     {
 24         return c-‘a‘;
 25     }
 26     void insert(char *s)
 27     {
 28         int u = 0,len = strlen(s);
 29         for (int i=0; i<len; ++i)
 30         {
 31             int c = idx(s[i]);
 32             if (!ch[u][c])
 33             {
 34                 ch[u][c] = ++size;
 35                 val[size] = 0;
 36                 memset(ch[size],0,sizeof(ch[size]));
 37             }
 38             u = ch[u][c];
 39         }
 40         val[u] ++;
 41     }
 42     void getfail()
 43     {
 44         queue<int>q;
 45         fail[0] = 0;
 46         for (int c=0; c<26; ++c) 
 47         {
 48             int u = ch[0][c];
 49             if (u) 
 50             {
 51                 fail[u] = 0;q.push(u);last[u] = 0;
 52             }
 53         }
 54         while (!q.empty())
 55         {
 56             int r = q.front();q.pop();
 57             for (int c=0; c<26; ++c)
 58             {
 59                 int u = ch[r][c];
 60                 if (!u)
 61                 {
 62                     ch[r][c] = ch[fail[r]][c];//有这一行就可以减去89行 
 63                     continue;
 64                 }
 65                 q.push(u);
 66                 int v = fail[r];
 67                 while (v && !ch[v][c]) v = fail[v];
 68                 fail[u] = ch[v][c];
 69                 last[u] = val[fail[u]] ? fail[u] : last[fail[u]];
 70             }
 71         }
 72     }
 73     void solve(int j)
 74     {
 75         if (!j) return ;
 76         if (val[j])
 77         {
 78             ret += val[j];
 79             val[j] = 0;
 80         }
 81         solve(last[j]);
 82     }
 83     void find(char *T)
 84     {
 85         int j = 0,len = strlen(T);
 86         for (int i=0; i<len; ++i)
 87         {
 88             int c = idx(T[i]);
 89         //    while (j && !ch[j][c]) j = fail[j];
 90             j = ch[j][c];
 91             if (val[j]) solve(j);
 92             else if (last[j]) solve(last[j]);
 93         }
 94     }
 95     
 96 }ac;
 97 
 98 int main()
 99 {
100     int t,n;
101     scanf("%d",&t);
102     while (t--) 
103     {
104         ac.clear();
105         scanf("%d",&n);
106         while (n--)
107         {
108             scanf("%s",s);
109             ac.insert(s);
110         }
111         ac.getfail();
112         scanf("%s",str);
113         ac.find(str);
114         printf("%d\n",ac.ret);
115     }
116     return 0;
117 }