HDOJ2222 Keywords Search-AC自动机

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.

现在才知道{scanf("%s",buf);ac.insert(buf);}和scanf("%s",buf),ac.insert(buf);不一样 TLE了十几次 一直在优化IO 以后要乖乖写分号了...

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<queue>
 using namespace std;
 ;
 ;
 struct AC{
     int size,root,next[maxn][maxc],fail[maxn],end[maxn];
     int newNode(){
         //memset(next[size],-1,sizeof(next[size]));
         ;i<maxc;i++) next[size][i]=-;
         end[size]=;
         ++size;
         ;
     }
     int idx(char c){
         return c-'a';
     }
     void init(){
         size=;
         //memset(end,0,sizeof(end));
         root=newNode();
     }
     void build(){
         queue<int> Q;
         fail[root]=root;
         ;i<maxc;i++){
             ) next[root][i]=root;
             else fail[next[root][i]]=root,Q.push(next[root][i]);
         }
         while(!Q.empty()){
             int u=Q.front();Q.pop();
             ;i<maxc;i++){
                 ) next[u][i]=next[fail[u]][i];
                 else fail[next[u][i]]=next[fail[u]][i],Q.push(next[u][i]);
             }
         }
     }
     int query(char* s){
         ;
         int l=strlen(s);
         ;i<l;i++){
             n=next[n][s[i]-'a'];
             int trav=n;
             while(trav!=root){
                 count+=end[trav];
                 end[trav]=;
                 trav=fail[trav];
             }
         }
         return count;
     }
     void insert(char* s){
         int n=root;
         int l=strlen(s);
         ;i<l;i++){
             ) next[n][s[i]-'a']=newNode();
             n=next[n][s[i]-'a'];
         }
         end[n]++;
     }
 };
 AC ac;
 *maxn];
 int main()
 {
     int T;scanf("%d",&T);
     while(T--){
         int n;scanf("%d",&n);
         ac.init();
         ;i<n;i++) {
             scanf("%s",buf);ac.insert(buf);
         }
         ac.build();
         scanf("%s",buf);
         printf("%d\n",ac.query(buf));
     }
     ;
 }