【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[0..2100000,'a'..'z']of longint;
    t:array[0..2100000]of record
                           s:longint;
                          end;
    root:array[0..500000]of longint;
    head,vet,next,dep:array[1..500000]of longint;
    f:array[1..550000,0..18]of longint;
    len:array[1..500000]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:=1 to 18 do
 begin
  if dep[u]<(1<<i) then break;
  f[u,i]:=f[f[u,i-1],i-1];
 end;
 e:=head[u];
 while e<>0 do
 begin
  v:=vet[e];
  if v<>f[u,0] then
  begin
   dep[v]:=dep[u]+1;
   f[v,0]:=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+1);
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*2);
 exit(query(map[a,zyd[i]],map[b,zyd[i]],map[c,zyd[i]],i+1));
end;

procedure dfs2(u:longint);
var e,v:longint;
begin
 e:=head[u];
 while e<>0 do
 begin
  v:=vet[e];
  if v<>f[u,0] then
  begin
   root[v]:=root[u];
   zyd:=len[e];
   ins(root[v],1);
   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:=0 to 18 do
  if d and (1<<i)>0 then x:=f[x,i];
 for i:=18 downto 0 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,0]);
end;

begin
 assign(input,'strings.in'); reset(input);
 assign(output,'strings.out'); rewrite(output);
 readln(n);
 for i:=1 to n-1 do
 begin
  readln(ch);
  j:=1; x:=0;
  while (ch[j]>='0')and(ch[j]<='9') do
  begin
   x:=x*10+ord(ch[j])-ord('0');
   inc(j);
  end;
  inc(j); y:=0;
  while (ch[j]>='0')and(ch[j]<='9') do
  begin
   y:=y*10+ord(ch[j])-ord('0');
   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(1);
 dfs2(1);
 readln(q);
 for i:=1 to q do
 begin
  readln(ch);
  j:=1; x:=0;
  while (ch[j]>='0')and(ch[j]<='9') do
  begin
   x:=x*10+ord(ch[j])-ord('0');
   inc(j);
  end;
  inc(j); y:=0;
  while (ch[j]>='0')and(ch[j]<='9') do
  begin
   y:=y*10+ord(ch[j])-ord('0');
   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],1));
 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[0..2100000,'a'..'z']of longint;
    t:array[0..2100000]of longint;
    root:array[1..500000]of longint;
    head,vet,next,dep,flag:array[1..500000]of longint;
    f:array[1..550000,0..18]of longint;
    len:array[1..500000]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:=0 to 18 do
  if d and (1<<i)>0 then x:=f[x,i];
 for i:=18 downto 0 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,0]);
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+1,map[p,h[i]]);
end;

procedure dfs(u:longint);
var e,v,i:longint;
begin
 for i:=1 to 18 do
 begin
  if dep[u]<(1<<i) then break;
  f[u,i]:=f[f[u,i-1],i-1];
 end;
 flag[u]:=1; e:=head[u];
 while e<>0 do
 begin
  v:=vet[e];
  if flag[v]=0 then
  begin
   dep[v]:=dep[u]+1;
   f[v,0]:=u;
   root[v]:=root[u];
   h:=len[e]; l:=length(h);
   build(1,root[v]);
   dfs(v);
  end;
  e:=next[e];
 end;
end;

function query(x,y,z,i:longint):longint;
begin
 if i=l+1 then exit(t[x]+t[y]-2*t[z]);
 exit(query(map[x,h[i]],map[y,h[i]],map[z,h[i]],i+1));
end;

begin
 assign(input,'bzoj4477.in'); reset(input);
 assign(output,'bzoj4477.out'); rewrite(output);
 readln(n);
 for i:=1 to n-1 do
 begin
  readln(ch);
  k:=length(ch); x:=0; y:=0; h:='';
  s:=1;
  for j:=1 to k do
  begin
   if ch[j]=' ' then begin inc(s); continue; end;
   if s=1 then x:=x*10+ord(ch[j])-ord('0');
   if s=2 then y:=y*10+ord(ch[j])-ord('0');
   if s=3 then h:=h+ch[j];
  end;
  add(x,y,h);
  add(y,x,h);
 end;
 dfs(1);
 readln(m);
 for i:=1 to m do
 begin
  readln(ch);
  k:=length(ch); x:=0; y:=0; s:=1; h:='';
  for j:=1 to k do
  begin
   if ch[j]=' ' then begin inc(s); continue; end;
   if s=1 then x:=x*10+ord(ch[j])-ord('0');
   if s=2 then y:=y*10+ord(ch[j])-ord('0');
   if s=3 then h:=h+ch[j];
  end;
  q:=lca(x,y);
  l:=length(h);
  writeln(query(root[x],root[y],root[q],1));
 end;
 close(input);
 close(output);
end.