HDU 2222

Keywords Search

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 33856 Accepted Submission(s): 10955

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

Author

Wiskey

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AC自动机

  1 #include <stdio.h>
  2 #include <queue>
  3 #include <cstring>
  4 #include <algorithm>
  5 using namespace std;
  6 const int maxn = 500010;
  7 int ch[maxn][26];
  8 char str[1000010];
  9 
 10 struct ACAutomation
 11 {
 12      int ch[maxn][26],fail[maxn],val[maxn],last[maxn],sz,root;
 13 
 14      int newnode()//新建一个节点
 15      {
 16          memset(ch[sz],0,sizeof(ch[sz]));//把这个节点的26个儿子置为0
 17          val[sz]=0;//单词结尾标志
 18          return sz++;
 19      }
 20 
 21      void init()
 22      {
 23          sz=0;
 24          root = newnode();
 25      }
 26 
 27      void insert()
 28      {
 29          int len = strlen(str);
 30          int now = root;
 31          for(int i=0;i<len;i++)//对每一位进行找到对应的节点
 32          {
 33              int &tmp = ch[now][str[i]-'a'];//now的儿子节点str[i]是否存在
 34              if(tmp==0)
 35                 tmp=newnode();//tmp=sz;
 36              now = tmp;
 37          }
 38          val[now]++;//单词结束标记
 39      }
 40 
 41      void getfail()
 42      {
 43          queue<int> q;
 44          fail[root] = root;//根节点的失败节点为它本身
 45           for(int i = 0;i < 26;i++)
 46           {
 47                 int u = ch[root][i];
 48                 if(u!=0)//根节点的子节点的失败节点为root
 49                 {
 50                     fail[u] = last[u] = 0;
 51                     q.push(u);
 52                 }
 53           }
 54             while(!q.empty())
 55             {
 56                 int now = q.front();
 57                 q.pop();
 58                 for(int i = 0;i < 26;i++)
 59                 {
 60                     int u = ch[now][i];//当前节点A
 61                     if(!u)
 62                         ch[now][i] = ch[fail[now]][i];//fail[now]就相当于我们说的父节点的失败节点C
 63                         //当前节点now没有i这个儿子u，直接让这个点的第i个儿子u指向now的失败节点的第i个儿子
 64                     else
 65                     {
 66                         fail[u] = ch[fail[now]][i];
 67                         //如果当前节点now有i这个儿子u，那么u的失败节点显然要指向C(即fail[now])的第i个儿子
 68                         last[u] = val[fail[u]] ? fail[u]:last[fail[u]];
 69                         //沿着u的失配指针走遇到的下一个单词节点的 节点的编号。
 70                         q.push(u);
 71                     }
 72                 }
 73             }
 74 
 75         }
 76 
 77         int query()
 78         {
 79             int len = strlen(str);
 80             int now = root;
 81             int ret = 0;
 82             for(int i = 0;i < len;i++)
 83             {
 84                 now = ch[now][str[i]-'a'];
 85                 int tmp = now;
 86                 while(tmp != root && val[tmp])
 87                 {
 88                     ret += val[tmp];
 89                     val[tmp] = 0;
 90                     //统计完了之后，记得修改标记，以免再次统计
 91                     tmp = last[tmp];
 92                     //对于在AC自动机走过的每一个单词节点，我们都要沿着后缀链接追踪，如果后缀链接指向的点是单词节点的话，这个点一定不能忘记统计。
 93                 }
 94             }
 95             return ret;
 96         }
 97 }ac;
 98 
 99 
100 int main(){
101     int t,n;
102     scanf("%d",&t);
103     while(t--){
104         ac.init();
105         scanf("%d",&n);
106         for(int i = 0;i < n;i++){
107             scanf("%s",str);
108             ac.insert();
109         }
110         ac.getfail();
111         scanf("%s",str);
112         printf("%d
",ac.query());
113     }
114     return 0;
115 }