[ac自动机]zoj3228 Searching the String

Searching the String

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Time Limit: 7 Seconds      Memory Limit: 129872 KB

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Little jay really hates to deal with string. But moondy likes it very much, and she's so mischievous that she often gives jay some dull problems related to string. And one day, moondy gave jay another problem, poor jay finally broke out and cried, " Who can help me? I'll bg him! "

So what is the problem this time?

First, moondy gave jay a very long string A. Then she gave him a sequence of very short substrings, and asked him to find how many times each substring appeared in string A. What's more, she would denote whether or not founded appearances of this substring are allowed to overlap.

At first, jay just read string A from begin to end to search all appearances of each given substring. But he soon felt exhausted and couldn't go on any more, so he gave up and broke out this time.

I know you're a good guy and will help with jay even without bg, won't you?

Input

Input consists of multiple cases( <= 20 ) and terminates with end of file.

For each case, the first line contains string A ( length <= 10^5 ). The second line contains an integer N ( N <= 10^5 ), which denotes the number of queries. The next N lines, each with an integer type and a string a ( length <= 6 ), type = 0 denotes substring a is allowed to overlap and type = 1 denotes not. Note that all input characters are lowercase.

There is a blank line between two consecutive cases.

Output

For each case, output the case number first ( based on 1 , see Samples ).

Then for each query, output an integer in a single line denoting the maximum times you can find the substring under certain rules.

Output an empty line after each case.

Sample Input

ab
2
0 ab
1 ab

abababac
2
0 aba
1 aba

abcdefghijklmnopqrstuvwxyz
3
0 abc
1 def
1 jmn

Sample Output

Case 1
1
1

Case 2
3
2

Case 3
1
1
0

Hint

In Case 2,you can find the first substring starting in position (indexed from 0) 0,2,4, since they're allowed to overlap. The second substring starts in position 0 and 4, since they're not allowed to overlap.

For C++ users, kindly use scanf to avoid TLE for huge inputs.

 

题目大意：给出一个字符串和若干个单词，问这些单词在字符串里面出现了多少次。单词前面为0表示这个单词可重叠出现，1为不可重叠出现。

分析：可重叠出现的单词可以直接用ac自动机就解决。至于不可重叠的单词，可以用一个数组来记录下每个这种单词上一次出现的位置，假如当前得到这个单词的位置和上一次的位置之间足够放下一个单词（即两个位置的坐标差大于等于单词长度）时才算一次，具体的可以看程序。我原先写的用数组实现的ac自动机，不过错了，然后用链表写，结果内存爆了..再用数组写了一次，ac了..这个程序都用了记录，应该很直观才对...

 

迅速完成ac自动机的论文先.....

 

 

codes：

type tnode=record n:array['a'..'z'] of longint; f,c:longint; end; var s:array[1..100000] of char; a:array[1..100000] of record typ,link:longint; s:string; end; t:array[0..600000] of tnode; b:array[1..1000000] of longint; ans:array[0..100000,0..1] of longint; pos:array[0..100000] of longint; n,len,k,root,tot:longint; procedure clr(x:longint); var i:char; begin for i:='a' to 'z' do t[x].n[i]:=-1; t[x].c:=0; t[x].f:=-1; end; procedure insert(s:string;num:longint); var i:longint; now:longint; begin now:=root; for i:=1 to length(s) do begin if t[now].n[s[i]]=-1 then begin inc(tot); t[now].n[s[i]]:=tot; clr(tot); end; now:=t[now].n[s[i]]; end; if t[now].c=0 then t[now].c:=num; a[num].link:=now; end; procedure build_ac; var temp,now:longint; i:char; head,tail:longint; begin head:=0; tail:=1; b[1]:=root; repeat inc(head); now:=b[head]; for i:='a' to 'z' do if t[now].n[i]<>-1 then begin temp:=t[now].f; while (temp<>-1) and (t[temp].n[i]=-1) do temp:=t[temp].f; if temp=-1 then t[t[now].n[i]].f:=root else t[t[now].n[i]].f:=t[temp].n[i]; inc(tail); b[tail]:=t[now].n[i]; end; until head>=tail; end; procedure init; var i:longint; c:char; begin len:=0; tot:=0; t[0].f:=-1; clr(0); read(c); while not seekeoln do begin inc(len); s[len]:=c; read(c); end; inc(len); s[len]:=c; readln(n); for i:=1 to n do begin readln(a[i].typ,a[i].s); while a[i].s[1]=' ' do delete(a[i].s,1,1); insert(a[i].s,i); end; readln; build_ac; end; procedure main; var i:longint; temp,now:longint; begin fillchar(ans,sizeof(ans),0); fillchar(pos,sizeof(pos),0); now:=root; for i:=1 to len do begin while (now<>-1) and (t[now].n[s[i]]=-1) do begin now:=now; now:=t[now].f; end; if now=-1 then now:=root else now:=t[now].n[s[i]]; temp:=now; while temp<>-1 do begin if t[temp].c>0 then begin inc(ans[t[temp].c,0]); if i-pos[t[temp].c]>=length(a[t[temp].c].s) then begin inc(ans[t[temp].c,1]); pos[t[temp].c]:=i; end; end; temp:=t[temp].f; end; end; for i:=1 to n do writeln(ans[t[a[i].link].c,a[i].typ]); end; begin root:=0; while not seekeof do begin inc(k); writeln('Case ',k); init; main; writeln; end; end.