[AC自动机]hdu2222 Keywords Search

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3406    Accepted Submission(s): 1177

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

15shehesayshrheryasherhs

 

Sample Output

3

 

Author

Wiskey

 

Recommend

lcy

 

题目大意：给出n个词，问在一个字符串中出现了多少个。

分析：一个ac自动机就解决了。但因为是第一次写，加上本题有个很容易tle的地方：用pascal不能用ansistring来处理，不然tle。结果，浪费了很多时间- -改成字符数组后，忘了数据规模是1000000，wa了很久= =茫然了我...

 

codes：

type tnode=^node; node=record next:array['a'..'z'] of tnode; count:longint; faild:tnode; end;var root:tnode; b:array[1..600001] of tnode; s:array[0..1000001] of char; k,n,ans:longint;procedure clr(t:tnode); var c:char; begin for c:='a' to 'z' do t^.next[c]:=nil; t^.count:=0; end;procedure insert(word:string;now:tnode); var i:longint; begin for i:=1 to length(word) do begin if now^.next[word[i]]=nil then begin new(now^.next[word[i]]); clr(now^.next[word[i]]); end; now:=now^.next[word[i]]; end; inc(now^.count); end;procedure build; var head,tail:longint; now,temp:tnode; i:char; begin head:=0; tail:=1; b[1]:=root; repeat inc(head); now:=b[head]; for i:='a' to 'z' do if now^.next[i]<>nil then begin if now=root then now^.next[i]^.faild:=root else begin temp:=now^.faild; while (temp<>nil) and (temp^.next[i]=nil)do temp:=temp^.faild; if temp=nil then now^.next[i]^.faild:=root else now^.next[i]^.faild:=temp^.next[i]; end; inc(tail); b[tail]:=now^.next[i]; end; until head>=tail; end;procedure init; var i:longint; word:string; begin readln(n); new(root); clr(root); root^.faild:=nil; for i:=1 to n do begin readln(word); insert(word,root); end; build; end;procedure cal(now:tnode); begin while (now<>nil) and (now^.count<>-1) do begin if now^.count>0 then inc(ans,now^.count); now^.count:=-1; now:=now^.faild; end; end;procedure main; var i,len:longint; ch:char; now:tnode; begin read(ch); len:=0; while ch in ['a'..'z'] do begin inc(len); s[len]:=ch; read(ch); end; now:=root; ans:=0; for i:=1 to len do begin while (now^.next[s[i]]=nil) and (now<>root) do now:=now^.faild; now:=now^.next[s[i]]; if now=nil then now:=root; cal(now); end; writeln(ans); end;begin readln(k); while k>0 do begin dec(k); init; main; end;end.