bzoj4390: [Usaco2015 dec]Max Flow(LCA+树上差分)

      题目大意：给出一棵树，n(n<=5w)个节点，k(k<=10w)次修改，每次给定s和t，把s到t的路径上的点权+1，问k次操作后最大点权。

      对于每次修改，给s和t的点权+1，给lca(s,t)和lca(s,t)的父亲的点权-1，每一个点的权就是它与它的子树权和，实际上就是树上的差分，又涨姿势了。。。

代码如下：

uses math;
type
  point=^rec;
  rec=record
    data:longint;
    next:point;
  end;
var
  n,m,x,y,i,ans,fa,kk:longint;
  k:point;
  d,siz,sum:array[0..100000]of longint;
  a:array[0..100000]of point;
  p:array[0..100000,0..20]of longint;

procedure dfs(x,fa:longint);
var
  k:point;
  l:longint;
begin
  d[x]:=d[fa]+1;
  l:=trunc(ln(n)/ln(2));
  p[x,0]:=fa;
  for i:=1 to l do
  p[x,i]:=p[p[x,i-1],i-1];
  k:=a[x];
  while k<>nil do
  begin
    if k^.data<>fa then dfs(k^.data,x);
    k:=k^.next;
  end;
end;

function lca(x,y:longint):longint;
var
  t,l,i:longint;
begin
  if d[x]>d[y] then
  begin
    t:=x;x:=y;y:=t;
  end;
  if d[x]<d[y] then
  begin
    l:=trunc(ln(d[y]-d[x])/ln(2))+1;
    for i:=l downto 0 do
    if d[y]-(1<<i)>=d[x] then
    y:=p[y,i];
  end;
  if x<>y then
  begin
    l:=trunc(ln(d[y])/ln(2));
    for i:=l downto 0 do
    if p[x,i]<>p[y,i] then
    begin
      x:=p[x,i];y:=p[y,i];
    end;
    x:=p[x,0];
  end;
  exit(x);
end;

procedure dfs2(x,fa:longint);
var
  k:point;
begin
  siz[x]:=sum[x];
  k:=a[x];
  while k<>nil do
  begin
    if k^.data<>fa then
    begin
      dfs2(k^.data,x);
      inc(siz[x],siz[k^.data]);
    end;
    k:=k^.next;
  end;
  ans:=max(ans,siz[x]);
end;

begin
  readln(n,kk);
  for i:=1 to n-1 do
  begin
    readln(x,y);
    new(k);
    k^.data:=y;
    k^.next:=a[x];
    a[x]:=k;
    new(k);
    k^.data:=x;
    k^.next:=a[y];
    a[y]:=k;
  end;
  dfs(1,0);
  for i:=1 to kk do
  begin
    readln(x,y);
    fa:=lca(x,y);
    inc(sum[x]);inc(sum[y]);dec(sum[fa]);
    if fa<>1 then dec(sum[p[fa,0]]);
  end;
  dfs2(1,0);
  writeln(ans);
end.