题意：给定一棵N个点树，询问这个树里面每个点到树上其他点的最大距离。
n<=10000
思路：设f[u,1],f[u,2]为以U为根向下的最长与次长，g[u,1],g[u,2]为从哪个儿子转移来
第一次dfs用V更新U，第二次dfs用U更新V，因为有V向U往上走的情况，这样做就可以处理了
可以发现这些数值中取最大值就是树的直径了
var f,g:array[..,..]of longint;
head,vet,next,len:array[..]of longint;
n,x,y,i,tot:longint; procedure add(a,b,c:longint);
begin
inc(tot);
next[tot]:=head[a];
vet[tot]:=b;
len[tot]:=c;
head[a]:=tot;
end; procedure dfs(u,fa:longint);
var e,v,t:longint;
begin
e:=head[u];
while e<> do
begin
v:=vet[e];
if v<>fa then
begin
dfs(v,u);
t:=f[v,]+len[e];
if t>f[u,] then
begin
f[u,]:=f[u,]; g[u,]:=g[u,];
f[u,]:=t; g[u,]:=v;
end
else if t>f[u,] then
begin
f[u,]:=t; g[u,]:=v;
end;
end;
e:=next[e];
end;
end; procedure dfs2(u,fa:longint);
var e,v,t:longint;
begin
e:=head[u];
while e<> do
begin
v:=vet[e];
if v<>fa then
begin
t:=f[u,]+len[e];
if v=g[u,] then t:=f[u,]+len[e];
if t>f[v,] then
begin
f[v,]:=f[v,]; g[v,]:=g[v,];
f[v,]:=t; g[v,]:=u;
end
else if t>f[v,] then
begin
f[v,]:=t; g[v,]:=u;
end;
dfs2(v,u);
end;
e:=next[e];
end;
end; begin
assign(input,'2196.in'); reset(input);
assign(output,'2196.out'); rewrite(output);
while not eof do
begin
fillchar(head,sizeof(head),);
tot:=;
fillchar(f,sizeof(f),);
fillchar(g,sizeof(g),);
read(n);
if n= then break;
for i:= to n do
begin
read(x,y);
add(i,x,y);
add(x,i,y);
end;
dfs(,-);
dfs2(,-);
for i:= to n do writeln(f[i,]);
end;
close(input);
close(output);
end.