bzoj3926

题目的意思是叶子不超过20个……听说当初zjoi不少人被坑

分别对每个叶子以它为根dfs出20个dfs树，这样整个树的任何一个子串，都是某个dfs树上一个点到它的一个子孙的路径

每个dfs树，根到叶子相当于一个串，这样相当于统计20*19个串的不同字串数目

这样我们很容易想到建立多串SAM（广义后缀树）统计不同的子串数即可

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

 var go:array[..,..] of longint;
     fa,mx:array[..] of longint;
     c,d,p:array[..] of longint;
     e:array[..] of node;
     v:array[..] of boolean;
     t,i,len,x,y,n,m:longint;
     ans:int64;

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

 function work(c,last:longint):longint;
   var np,q,p,nq:longint;
   begin
     p:=last;
     if go[p,c]= then
     begin
       inc(t); work:=t; np:=t;
       mx[np]:=mx[p]+;
       while (p>) and (go[p,c]=) do
       begin
         go[p,c]:=np;
         p:=fa[p];
       end;
       if p= then fa[np]:=
       else begin
         q:=go[p,c];
         if mx[q]=mx[p]+ then fa[np]:=q
         else begin
           inc(t); nq:=t;
           mx[nq]:=mx[p]+;
           go[nq]:=go[q];
           fa[nq]:=fa[q];
           fa[q]:=nq; fa[np]:=nq;
           while go[p,c]=q do
           begin
             go[p,c]:=nq;
             p:=fa[p];
           end;
         end;
       end;
     end
     else begin
       q:=go[p,c];
       if mx[q]=mx[p]+ then exit(q)
       else begin
         inc(t); nq:=t;
         mx[nq]:=mx[p]+;
         go[nq]:=go[q];
         fa[nq]:=fa[q];
         fa[q]:=nq;
         while go[p,c]=q do
         begin
           go[p,c]:=nq;
           p:=fa[p];
         end;
       end;
       exit(go[last,c]);
     end;
   end;

 procedure dfs(x,last:longint);
   var i,y:longint;
   begin
     last:=work(c[x],last);
     v[x]:=true;
     i:=p[x];
     while i<> do
     begin
       y:=e[i].po;
       if not v[y] then dfs(y,last);
       i:=e[i].next;
     end;
   end;

 begin
   readln(n,m);
   for i:= to n do
     read(c[i]);
   for i:= to n- do
   begin
     readln(x,y);
     add(x,y);
     add(y,x);
   end;
   t:=;
   for i:= to n do
     if d[i]= then
     begin
       fillchar(v,sizeof(v),false);
       dfs(i,);
     end;
   for i:= to t do
     ans:=ans+mx[i]-mx[fa[i]];
   writeln(ans);
 end.