題意：n個點被n-1條邊連接成了一顆樹，給出a~b和c~d兩個區間，

表示點的標號請你求出兩個區間內各選一點之間的最大距離，即你需要求出max｛dis(i,j) |a<=i<=b,c<=j<=d｝

n<=100000 len[i]<=100000

思路：兩年前張老師出的模擬賽裏的題

設區間[a,b]中最大距離點對為[x1,y1]

[c,d]為[x2,y2]

則兩區間各取一個點的最值只有兩兩組合4種可能

而線段樹中pushup有6種，可以在同一個區間中取兩點

求LCA需要LCA轉RMQ O(1)做 倍增或剖的話都會T

然並卵 P的常數太大了 早日轉C保平安

  1 type arr=record
 

  2           a,b:longint;
  3          end;
  4 var f,g:array[1..210000,0..20]of longint;
  5     t:array[1..1000000]of arr;
  6     a,b,st,dep,dis,head,vet,next,len:array[1..300000]of longint;
  7     n,m,i,tot,x,y,z,ans,time,que,j,x1,y1,x2,y2,tmp,tx,ty:longint;
  8     p,q:arr;
  9 
 10 procedure add(a,b,c:longint);
 
 11 begin
 12  inc(tot);
 13  next[tot]:=head[a];
 14  vet[tot]:=b;
 15  len[tot]:=c;
 16  head[a]:=tot;
 17 end;
 18 
 19 function max(x,y:longint):longint;
 20 begin
 21  if x>y then exit(x);
 22  exit(y);
 23 end;
 24 
 25 procedure dfs(u,fa:longint);
 26 var e,v:longint;
 27 begin
 28  inc(time); st[u]:=time; a[time]:=dep[u]; b[time]:=u;
 
 29  e:=head[u];
 30  while e<>0 do
 31  begin
 32   v:=vet[e];
 33   if v<>fa then
 34   begin
 35    dep[v]:=dep[u]+1;
 36    dis[v]:=dis[u]+len[e];
 37    dfs(v,u);
 38    inc(time); a[time]:=dep[u]; b[time]:=u;
 39   end;
 40   e:=next[e];
 41  end;
 42 end;
 43 
 44 function min(x,y:longint):longint;
 45 begin
 46  if x<y then exit(x);
 47  exit(y);
 48 end;
 49 
 50 function lca(x,y:longint):longint;
 51 var len,l,x1,y1:longint;
 52 begin
 53  x1:=st[x]; y1:=st[y];
 54  if x1>y1 then begin x1:=st[y]; y1:=st[x]; end;
 55  len:=y1-x1+1;
 56  l:=trunc(ln(len)/ln(2));
 57  if f[x1,l]<f[y1-(1<<l)+1,l] then exit(g[x1,l])
 58   else exit(g[y1-(1<<l)+1,l]);
 59 end;
 60 
 61 function clac(x,y:longint):longint;
 62 var q:longint;
 63 begin
 64  q:=lca(x,y);
 65  exit(dis[x]+dis[y]-2*dis[q]);
 66 end;
 67 
 68 function pushup(x,y:arr):arr;
 69 var t:arr;
 70 begin
 71  t:=x;
 72  if clac(y.a,y.b)>clac(t.a,t.b) then t:=y;
 73  if clac(x.a,y.a)>clac(t.a,t.b) then
 74  begin
 75   t.a:=x.a; t.b:=y.a;
 76  end;
 77  if clac(x.a,y.b)>clac(t.a,t.b) then
 78  begin
 79   t.a:=x.a; t.b:=y.b;
 80  end;
 81  if clac(x.b,y.a)>clac(t.a,t.b) then
 82  begin
 83   t.a:=x.b; t.b:=y.a;
 84  end;
 85  if clac(x.b,y.b)>clac(t.a,t.b) then
 86  begin
 87   t.a:=x.b; t.b:=y.b;
 88  end;
 89  exit(t);
 90 end;
 91 
 92 procedure build(l,r,p:longint);
 93 var mid:longint;
 94 begin
 95  if l=r then
 96  begin
 97   t[p].a:=l; t[p].b:=l;
 98   exit;
 99  end;
100  mid:=(l+r)>>1;
101  build(l,mid,p<<1);
102  build(mid+1,r,p<<1+1);
103  t[p]:=pushup(t[p<<1],t[p<<1+1]);
104 end;
105 
106 function query(l,r,x,y,p:longint):arr;
107 var mid:longint;
108 begin
109  if (l>=x)and(r<=y) then exit(t[p]);
110  mid:=(l+r)>>1;
111  query.a:=x; query.b:=x;
112  if x<=mid then query:=pushup(query,query(l,mid,x,y,p<<1));
113  if y>mid then query:=pushup(query,query(mid+1,r,x,y,p<<1+1));
114 end;
115 
116 begin
117  assign(input,‘51nod1766.in‘); reset(input);
118  assign(output,‘51nod1766.out‘); rewrite(output);
119  readln(n);
120  for i:=1 to n-1 do
121  begin
122   readln(x,y,z);
123   add(x,y,z);
124   add(y,x,z);
125  end;
126  dfs(1,0);
127  for i:=1 to time do
128  begin
129   f[i,0]:=a[i]; g[i,0]:=b[i];
130  end;
131  m:=trunc(ln(time)/ln(2));
132  for i:=1 to m do
133   for j:=1 to time-(1<<i)+1 do
134    if f[j,i-1]<f[j+(1<<(i-1)),i-1] then
135    begin
136     f[j,i]:=f[j,i-1]; g[j,i]:=g[j,i-1];
137    end
138     else
139     begin
140      f[j,i]:=f[j+(1<<(i-1)),i-1];
141      g[j,i]:=g[j+(1<<(i-1)),i-1];
142     end;
143  build(1,n,1);
144  readln(que);
145  for i:=1 to que do
146  begin
147   readln(x1,y1,x2,y2);
148   p:=query(1,n,x1,y1,1);
149   q:=query(1,n,x2,y2,1);
150   ans:=0;
151   ans:=max(ans,clac(p.a,q.a));
152   ans:=max(ans,clac(p.a,q.b));
153   ans:=max(ans,clac(p.b,q.a));
154   ans:=max(ans,clac(p.b,q.b));
155   writeln(ans);
156  end;
157  close(input);
158  close(output);
159 end.

【51NOD1766】樹上的最遠點對（線段樹，LCA，RMQ）