bzoj3172

这里学习AC自动机
其实对KMP和trie掌握好了之后很容易扩展成AC自动机的
这里运用了一个性质
由失配指针反向可以建成一棵fail树
x串在y串中的出现的次数即为在fail树上以x结尾节点为根的子树中有多少个节点属于y串，这很好理解
而推广到这题，在文章中出现的次数即为单词结尾节点为根的子树的规模
注意这道题有相同的单词
当初这题我是用后缀数组做的O(LlogL)
而自动机复杂度为O(L),快了很多

 type node=record
        po,next:longint;
      end;

 var e:array[..] of node;
     trie:array[..,'a'..'z'] of longint;
     q,f,p,w:array[..] of longint;
     v:array[..] of longint;
     t,i,j,k,n,l,len:longint;
     s:ansistring;

 procedure add(x,y:longint);
   begin
     inc(len);
     e[len].po:=y;
     e[len].next:=p[x];
     p[x]:=len;
   end;

 procedure bfs;
   var x,y,h,r,j:longint;
       c:char;
   begin
     h:=;
     r:=;
     for c:='a' to 'z' do
       if trie[,c]> then
       begin
         add(,trie[,c]);
         inc(r);
         q[r]:=trie[,c];
       end;

     while h<=r do
     begin
       x:=q[h];
       for c:='a' to 'z' do
         if trie[x,c]> then
         begin
           y:=trie[x,c];
           inc(r);
           q[r]:=y;
           j:=f[x];
           while (j>) and (trie[j,c]=) do j:=f[j];
           f[y]:=trie[j,c];
           add(trie[j,c],y);
         end;
       inc(h);
     end;
   end;

 procedure dfs(x:longint);
   var i,y:longint;
   begin
     i:=p[x];
     while i<> do
     begin
       y:=e[i].po;
       dfs(y);
       w[x]:=w[y]+w[x];
       i:=e[i].next;
     end;
   end;

 begin
   readln(n);
   for i:= to n do
   begin
     readln(s);
     j:=;
     l:=length(s);
     for k:= to l do
     begin
       if trie[j,s[k]]= then
       begin
         inc(t);
         trie[j,s[k]]:=t;
       end;
       j:=trie[j,s[k]];
       inc(w[j]);
     end;
     v[i]:=j;
   end;
   bfs;
   dfs();
   for i:= to n do
     writeln(w[v[i]]);
 end.

ac自动机

 var h,rank,x,y,sum,sa:array[..] of longint;
     f:array[..,..] of longint;
     where,len,d:array[..] of longint;
     now, k,ans,i,j,p,n,m,t,l,r:longint;
     s,cc:ansistring;
 function min(a,b:longint):longint;
   begin
     if a>b then exit(b) else exit(a);
   end;

 function lcp(a,b:longint):longint;
   var k:longint;
   begin
     k:=trunc(ln(b-a+)/ln());
     exit(min(f[a,k],f[b-d[k]+,k]));
   end;

 begin
   readln(t);
   for i:= to t do
   begin
     readln(cc);
     where[i]:=length(s)+;
     len[i]:=length(cc);
     s:=s+cc;
     if i<>t then s:=s+' ';
   end;
 {  for i:=1 to t do
      writeln(where[i],' ',len[i]);}
   n:=length(s);
   for i:= to n do
   begin
     y[i]:=ord(s[i]);
     inc(sum[y[i]]);
   end;
   m:=;
   for i:= to  do
     inc(sum[i],sum[i-]);
   for i:= to n do
   begin
     sa[sum[y[i]]]:=i;
     dec(sum[y[i]]);
   end;
   p:=;
   rank[sa[]]:=;
   for i:= to n do
   begin
     if y[sa[i]]<>y[sa[i-]] then inc(p);
     rank[sa[i]]:=p;
   end;
   m:=p;
   j:=;
   while m<n do
   begin
     y:=rank;
     fillchar(sum,sizeof(sum),);
     p:=;
     for i:=n-j+ to n do
     begin
       inc(p);
       x[p]:=i;
     end;
     for i:= to n do
       if sa[i]>j then
       begin
         inc(p);
         x[p]:=sa[i]-j;
       end;
     for i:= to n do
     begin
       rank[i]:=y[x[i]];
       inc(sum[rank[i]]);
     end;
     for i:= to m do
       inc(sum[i],sum[i-]);
     for i:=n downto  do
     begin
       sa[sum[rank[i]]]:=x[i];
       dec(sum[rank[i]]);
     end;
     p:=;
     rank[sa[]]:=;
     for i:= to n do
     begin
       if (y[sa[i]]<>y[sa[i-]]) or (y[sa[i]+j]<>y[sa[i-]+j]) then inc(p);
       rank[sa[i]]:=p;
     end;
     j:=j shl ;
     m:=p;
   end;
   h[]:=;
   p:=;
   for i:= to n do
   begin
     if rank[i]= then continue;
     j:=sa[rank[i]-];
     while s[i+p]=s[j+p] do inc(p);
     h[rank[i]]:=p;
     if p> then dec(p);
   end;
   k:=trunc(ln(n)/ln());
   d[]:=;
   for i:= to k do
     d[i]:=d[i-]*;
   for i:= to n do
     f[i,]:=h[i];
   for j:= to k do
     for i:= to n do
       if i+d[j]-<=n then
       begin
         f[i,j]:=min(f[i,j-],f[i+d[j-],j-]);
       end
       else break;
   for i:= to t do
   begin
     ans:=;
     k:=-;
     l:=;
     now:=rank[where[i]];
     r:=now-;
     while l<=r do
     begin
       m:=(l+r) shr ;
       if lcp(m+,now)>=len[i] then
       begin
         k:=m;
         r:=m-
       end
       else l:=m+;
     end;
     if k<>- then ans:=ans+now-k;
     l:=now+;
     k:=-;
     r:=n;
     while l<=r do
     begin
       m:=(l+r) shr ;
       if lcp(now+,m)>=len[i] then
       begin
         k:=m;
         l:=m+;
       end
       else r:=m-;
     end;
     if k<>- then ans:=ans+k-now;
     writeln(ans);
   end;
 end.

后缀数组