【51NOD1766】树上的最远点对（线段树，LCA，RMQ）

题意：n个点被n-1条边连接成了一颗树，给出a~b和c~d两个区间，

表示点的标号请你求出两个区间内各选一点之间的最大距离，即你需要求出max｛dis(i,j) |a<=i<=b,c<=j<=d｝

n<=100000 len[i]<=100000

思路：两年前张老师出的模拟赛里的题

设区间[a,b]中最大距离点对为[x1,y1]

[c,d]为[x2,y2]

则两区间各取一个点的最值只有两两组合4种可能

而线段树中pushup有6种，可以在同一个区间中取两点

求LCA需要LCA转RMQ O(1)做 倍增或剖的话都会T

然并卵 P的常数太大了 早日转C保平安

 type arr=record
           a,b:longint;
          end;
 var f,g:array[..,..]of longint;
     t:array[..]of arr;
     a,b,st,dep,dis,head,vet,next,len:array[..]of longint;
     n,m,i,tot,x,y,z,ans,time,que,j,x1,y1,x2,y2,tmp,tx,ty:longint;
     p,q:arr;

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

 function max(x,y:longint):longint;
 begin
  if x>y then exit(x);
  exit(y);
 end;

 procedure dfs(u,fa:longint);
 var e,v:longint;
 begin
  inc(time); st[u]:=time; a[time]:=dep[u]; b[time]:=u;
  e:=head[u];
  while e<> do
  begin
   v:=vet[e];
   if v<>fa then
   begin
    dep[v]:=dep[u]+;
    dis[v]:=dis[u]+len[e];
    dfs(v,u);
    inc(time); a[time]:=dep[u]; b[time]:=u;
   end;
   e:=next[e];
  end;
 end;

 function min(x,y:longint):longint;
 begin
  if x<y then exit(x);
  exit(y);
 end;

 function lca(x,y:longint):longint;
 var len,l,x1,y1:longint;
 begin
  x1:=st[x]; y1:=st[y];
  if x1>y1 then begin x1:=st[y]; y1:=st[x]; end;
  len:=y1-x1+;
  l:=trunc(ln(len)/ln());
  if f[x1,l]<f[y1-(<<l)+,l] then exit(g[x1,l])
   else exit(g[y1-(<<l)+,l]);
 end;

 function clac(x,y:longint):longint;
 var q:longint;
 begin
  q:=lca(x,y);
  exit(dis[x]+dis[y]-*dis[q]);
 end;

 function pushup(x,y:arr):arr;
 var t:arr;
 begin
  t:=x;
  if clac(y.a,y.b)>clac(t.a,t.b) then t:=y;
  if clac(x.a,y.a)>clac(t.a,t.b) then
  begin
   t.a:=x.a; t.b:=y.a;
  end;
  if clac(x.a,y.b)>clac(t.a,t.b) then
  begin
   t.a:=x.a; t.b:=y.b;
  end;
  if clac(x.b,y.a)>clac(t.a,t.b) then
  begin
   t.a:=x.b; t.b:=y.a;
  end;
  if clac(x.b,y.b)>clac(t.a,t.b) then
  begin
   t.a:=x.b; t.b:=y.b;
  end;
  exit(t);
 end;

 procedure build(l,r,p:longint);
 var mid:longint;
 begin
  if l=r then
  begin
   t[p].a:=l; t[p].b:=l;
   exit;
  end;
  mid:=(l+r)>>;
  build(l,mid,p<<);
  build(mid+,r,p<<+);
  t[p]:=pushup(t[p<<],t[p<<+]);
 end;

 function query(l,r,x,y,p:longint):arr;
 var mid:longint;
 begin
  if (l>=x)and(r<=y) then exit(t[p]);
  mid:=(l+r)>>;
  query.a:=x; query.b:=x;
  if x<=mid then query:=pushup(query,query(l,mid,x,y,p<<));
  if y>mid then query:=pushup(query,query(mid+,r,x,y,p<<+));
 end;

 begin
  assign(input,'51nod1766.in'); reset(input);
  assign(output,'51nod1766.out'); rewrite(output);
  readln(n);
  for i:= to n- do
  begin
   readln(x,y,z);
   add(x,y,z);
   add(y,x,z);
  end;
  dfs(,);
  for i:= to time do
  begin
   f[i,]:=a[i]; g[i,]:=b[i];
  end;
  m:=trunc(ln(time)/ln());
  for i:= to m do
   for j:= to time-(<<i)+ do
    if f[j,i-]<f[j+(<<(i-)),i-] then
    begin
     f[j,i]:=f[j,i-]; g[j,i]:=g[j,i-];
    end
     else
     begin
      f[j,i]:=f[j+(<<(i-)),i-];
      g[j,i]:=g[j+(<<(i-)),i-];
     end;
  build(,n,);
  readln(que);
  for i:= to que do
  begin
   readln(x1,y1,x2,y2);
   p:=query(,n,x1,y1,);
   q:=query(,n,x2,y2,);
   ans:=;
   ans:=max(ans,clac(p.a,q.a));
   ans:=max(ans,clac(p.a,q.b));
   ans:=max(ans,clac(p.b,q.a));
   ans:=max(ans,clac(p.b,q.b));
   writeln(ans);
  end;
  close(input);
  close(output);
 end.