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.