【BZOJ4477】字符串树（可持久化Trie）

此题花费我整整三天的功夫。还在NoiP贴吧发过贴。

最后发现trie树建新节点时信息未完全复制，真是愚蠢之极。

言归正传。

如果我们已经知道了每个点上的trie树那么询问就是sum[x]+sum[y]-sum[lca(x,y)]*2

然后就是trie树变可持久化。

DFS2中插入所有字符串，建立新节点，复制出现次数与trie树的next指针。

然后就没有然后了。

  map:array[..,'a'..'z']of longint;
     t:array[..]of record
                            s:longint;
                           end;
     root:array[..]of longint;
     head,vet,next,dep:array[..]of longint;
     f:array[..,..]of longint;
     len:array[..]of string;
     n,i,x,y,tot,tmp,cnt,j,k,q:longint;
     zyd,ch1,ch:string;

 procedure add(a,b:longint;c:string);
 begin
  inc(tot);
  next[tot]:=head[a];
  vet[tot]:=b;
  len[tot]:=c;
  head[a]:=tot;
 end;

 procedure dfs1(u:longint);
 var e,v,i:longint;
 begin
  for i:= to  do
  begin
   if dep[u]<(<<i) then break;
   f[u,i]:=f[f[u,i-],i-];
  end;
  e:=head[u];
  while e<> do
  begin
   v:=vet[e];
   if v<>f[u,] then
   begin
    dep[v]:=dep[u]+;
    f[v,]:=u;
    dfs1(v);
   end;
   e:=next[e];
  end;
 end;

 procedure ins(var x:longint;i:longint);
 begin
  inc(cnt); t[cnt]:=t[x];
  map[cnt]:=map[x];
  x:=cnt; inc(t[x].s);
  if i<=length(zyd) then ins(map[x,zyd[i]],i+);
 end;

 function query(a,b,c,i:longint):longint;
 begin
 // writeln(a,' ',b,' ',c);
  if i>length(zyd) then exit(t[a].s+t[b].s-t[c].s*);
  exit(query(map[a,zyd[i]],map[b,zyd[i]],map[c,zyd[i]],i+));
 end;

 procedure dfs2(u:longint);
 var e,v:longint;
 begin
  e:=head[u];
  while e<> do
  begin
   v:=vet[e];
   if v<>f[u,] then
   begin
    root[v]:=root[u];
    zyd:=len[e];
    ins(root[v],);
    dfs2(v);
   end;
   e:=next[e];
  end;
 end;

 procedure swap(var x,y:longint);
 var t:longint;
 begin
  t:=x; x:=y; y:=t;
 end;

 function lca(x,y:longint):longint;
 var i,d:longint;
 begin
  if dep[x]<dep[y] then swap(x,y);
  d:=dep[x]-dep[y];
  for i:= to  do
   if d and (<<i)> then x:=f[x,i];
  for i:= downto  do
   if f[x,i]<>f[y,i] then
   begin
    x:=f[x,i]; y:=f[y,i];
   end;
  if x=y then exit(x);
  exit(f[x,]);
 end;

 begin
  assign(input,'strings.in'); reset(input);
  assign(output,'strings.out'); rewrite(output);
  readln(n);
  for i:= to n- do
  begin
   readln(ch);
   j:=; x:=;
   while (ch[j]>='')and(ch[j]<='') do
   begin
    x:=x*+ord(ch[j])-ord('');
    inc(j);
   end;
   inc(j); y:=;
   while (ch[j]>='')and(ch[j]<='') do
   begin
    y:=y*+ord(ch[j])-ord('');
    inc(j);
   end;
   inc(j);
   ch1:='';
   for k:=j to length(ch) do ch1:=ch1+ch[k];
   add(x,y,ch1);
   add(y,x,ch1);
  end;

  //f[1,0]:=1;

  dfs1();
  dfs2();
  readln(q);
  for i:= to q do
  begin
   readln(ch);
   j:=; x:=;
   while (ch[j]>='')and(ch[j]<='') do
   begin
    x:=x*+ord(ch[j])-ord('');
    inc(j);
   end;
   inc(j); y:=;
   while (ch[j]>='')and(ch[j]<='') do
   begin
    y:=y*+ord(ch[j])-ord('');
    inc(j);
   end;
   inc(j);
   ch1:='';
   for k:=j to length(ch) do ch1:=ch1+ch[k];
   tmp:=lca(x,y);
   zyd:=ch1;
   writeln(query(root[x],root[y],root[tmp],));
  end;

  close(input);
  close(output);
 end.

2016.12.25

学会主席树后重写了一遍，感觉异常轻松

因为此题询问的是边上的信息，所以LCA点不会被重复计算，询问时也就不用-1

如果是点则要-1

 var map:array[..,'a'..'z']of longint;
     t:array[..]of longint;
     root:array[..]of longint;
     head,vet,next,dep,flag:array[..]of longint;
     f:array[..,..]of longint;
     len:array[..]of string;
     n,i,x,y,tot,tmp,cnt,j,k,s,m,l,q:longint;
     ch,h:string;

 procedure swap(var x,y:longint);
 var t:longint;
 begin
  t:=x; x:=y; y:=t;
 end;

 procedure add(a,b:longint;c:string);
 begin
  inc(tot);
  next[tot]:=head[a];
  vet[tot]:=b;
  len[tot]:=c;
  head[a]:=tot;
 end;

 function lca(x,y:longint):longint;
 var d,i:longint;
 begin
  if dep[x]<dep[y] then swap(x,y);
  d:=dep[x]-dep[y];
  for i:= to  do
   if d and (<<i)> then x:=f[x,i];
  for i:= downto  do
   if f[x,i]<>f[y,i] then
   begin
    x:=f[x,i]; y:=f[y,i];
   end;
  if x=y then exit(x);
  exit(f[x,]);
 end;

 procedure build(i:longint;var p:longint);
 begin
  inc(cnt); t[cnt]:=t[p];
  map[cnt]:=map[p];
  p:=cnt; inc(t[p]);
  if i<=l then build(i+,map[p,h[i]]);
 end;

 procedure dfs(u:longint);
 var e,v,i:longint;
 begin
  for i:= to  do
  begin
   if dep[u]<(<<i) then break;
   f[u,i]:=f[f[u,i-],i-];
  end;
  flag[u]:=; e:=head[u];
  while e<> do
  begin
   v:=vet[e];
   if flag[v]= then
   begin
    dep[v]:=dep[u]+;
    f[v,]:=u;
    root[v]:=root[u];
    h:=len[e]; l:=length(h);
    build(,root[v]);
    dfs(v);
   end;
   e:=next[e];
  end;
 end;

 function query(x,y,z,i:longint):longint;
 begin
  if i=l+ then exit(t[x]+t[y]-*t[z]);
  exit(query(map[x,h[i]],map[y,h[i]],map[z,h[i]],i+));
 end;

 begin
  assign(input,'bzoj4477.in'); reset(input);
  assign(output,'bzoj4477.out'); rewrite(output);
  readln(n);
  for i:= to n- do
  begin
   readln(ch);
   k:=length(ch); x:=; y:=; h:='';
   s:=;
   for j:= to k do
   begin
    if ch[j]=' ' then begin inc(s); continue; end;
    if s= then x:=x*+ord(ch[j])-ord('');
    if s= then y:=y*+ord(ch[j])-ord('');
    if s= then h:=h+ch[j];
   end;
   add(x,y,h);
   add(y,x,h);
  end;
  dfs();
  readln(m);
  for i:= to m do
  begin
   readln(ch);
   k:=length(ch); x:=; y:=; s:=; h:='';
   for j:= to k do
   begin
    if ch[j]=' ' then begin inc(s); continue; end;
    if s= then x:=x*+ord(ch[j])-ord('');
    if s= then y:=y*+ord(ch[j])-ord('');
    if s= then h:=h+ch[j];
   end;
   q:=lca(x,y);
   l:=length(h);
   writeln(query(root[x],root[y],root[q],));
  end;
  close(input);
  close(output);
 end.