題目連接：點我
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

ac自動機裸題

  1 #include<queue>
  2 #include<set>
  3 #include<cstdio>
  4 #include <iostream>
  5 #include<algorithm>
  6 #include<cstring>
  7
 
 #include<cmath>
  8 using namespace std;
  9 char c[1000000+10];
 10 struct acnode{
 11     int sum;
 12     acnode* next[26];
 13     acnode* fail;
 14     acnode(){
 15         for(int i =0;i<26;i++)
 16             next[i]=NULL;
 17         fail= NULL;
 18         sum=0;
 19     }
 20 };
 21 acnode *root;
 22 int cnt;
 23 acnode* newnode(){
 24     acnode *p = new acnode;
 25     for(int i =0;i<26;i++)
 26         p->next[i]=NULL;
 27     p->fail = NULL;
 28     p->sum=0;
 29     return p;
 30 }
 31 //插入函數
 32 void Insert(char *s)
 33 {
 34     acnode *p = root;
 35     for(int i = 0; s[i]; i++)
 36     {
 37         int x = s[i] - ‘a‘;
 38         if(p->next[x]==NULL)
 39         {
 40             acnode *nn=newnode();
 41             for(int j=0;j<26;j++)
 42                 nn->next[j] = NULL;
 43             nn->sum = 0;
 44             nn->fail = NULL;
 45             p->next[x]=nn;
 46         }
 47         p = p->next[x];
 48     }
 49     p->sum++;
 50 }
 51 //獲取fail指針，在插入結束之後使用
 52 void getfail(){
 53     queue<acnode*> q;
 54     for(int i = 0 ; i < 26 ; i ++ )
 55     {
 56         if(root->next[i]!=NULL){
 57             root->next[i]->fail = root;
 58             q.push(root->next[i]);
 59         }
 60     }
 61     while(!q.empty()){
 62         acnode* tem = q.front();
 63         q.pop();
 64         for(int i = 0;i<26;i++){
 65             if(tem->next[i]!=NULL)
 66             {
 67                 acnode *p;
 68                 if(tem == root){
 69                     tem->next[i]->fail = root;
 70                 }
 71                 else
 72                 {
 73                     p = tem->fail;
 74                     while(p!=NULL){
 75                         if(p->next[i]!=NULL){
 76                             tem->next[i]->fail = p->next[i];
 77                             break;
 78                         }
 79                         p=p->fail;
 80                     }
 81                     if(p==NULL)
 82                         tem->next[i]->fail = root;
 83                 }
 84                 q.push(tem->next[i]);
 85             }
 86         }
 87     }
 88 }
 89 //匹配函數
 90 void ac_automation(char *ch)
 91 {
 92     acnode *p = root;
 93     int len = strlen(ch);
 94     for(int i = 0; i < len; i++)
 95     {
 96         int x = ch[i] - ‘a‘;
 97         while(p->next[x]==NULL && p != root)//沒匹配到，那麽就找fail指針。
 98             p = p->fail;
 99         p = p->next[x];
100         if(!p)
101             p = root;
102         acnode *temp = p;
103         while(temp != root)
104         {
105            if(temp->sum >= 0)
106             /*
107             在這裏已經匹配成功了，執行想執行的操作即可，怎麽改看題目需求+
108             */
109            {
110                cnt += temp->sum;
111                temp->sum = -1;
112            }
113            else break;
114            temp = temp->fail;
115         }
116     }
117 }
118 
119 int main()
120 {   int t;
121     scanf("%d",&t);
122     while(t--){
123         cnt = 0;
124     int n;
125     cin>>n;
126 
127     root = newnode();
128     for(int i = 0 ;i < n;i++){
129         scanf("%s",c);
130         Insert(c);
131     }
132     getfail();
133     //int m ;
134     //cin>> m;
135     //for(int i = 0;i<m;i++){
136         scanf("%s",c);
137         ac_automation(c);
138     //}
139     cout<<cnt<<endl;
140     } 
141 
142     return 0;
143 }

杭電 2222 Keywords Search