bzoj3531

时间:2023-03-08 20:54:39

不难想到树链剖分
这题的难点是记录的是路径上宗教相同的点
裸的想法是对每一种宗教都开一棵线段树,记录每个点的评级
但显然这样会爆空间,仔细分析一下,这些线段树内很多点压根就没用到
因此我们考虑对线段树动态开点,不难发现每次修改最多要开线段树上O(2*logn)个点,是可以接受的
然后就是打码的问题了

 type node=record
po,next:longint;
end;
link=record
l,r,s,m:longint;
end; var tree:array[..*] of link;
fa,size,d,top,p,c,h,b,w:array[..] of longint;
e:array[..] of node;
len,t,i,n,q,x,y:longint;
ch:char; function max(a,b:longint):longint;
begin
if a>b then exit(a) else exit(b);
end; procedure swap(var a,b:longint);
var c:longint;
begin
c:=a;
a:=b;
b:=c;
end; procedure add(x,y:longint);
begin
inc(len);
e[len].po:=y;
e[len].next:=p[x];
p[x]:=len;
end; procedure update(x:longint);
begin
tree[x].s:=tree[tree[x].l].s+tree[tree[x].r].s;
tree[x].m:=max(tree[tree[x].l].m,tree[tree[x].r].m);
end; procedure dfs1(x:longint);
var i,y:longint;
begin
size[x]:=;
i:=p[x];
while i<> do
begin
y:=e[i].po;
if fa[x]<>y then
begin
d[y]:=d[x]+;
fa[y]:=x;
dfs1(y);
size[x]:=size[x]+size[y];
end;
i:=e[i].next;
end;
end; procedure dfs2(x:longint);
var i,y,q:longint;
begin
q:=;
inc(t);
c[x]:=t;
i:=p[x];
while i<> do
begin
y:=e[i].po;
if (c[y]=) and (size[y]>size[q]) then q:=y;
i:=e[i].next;
end;
if q<> then
begin
top[q]:=top[x];
dfs2(q);
end;
i:=p[x];
while i<> do
begin
y:=e[i].po;
if c[y]= then
begin
top[y]:=y;
dfs2(y);
end;
i:=e[i].next;
end;
end; function change(last,l,r,x,y:longint):longint;
var m,q:longint;
begin
inc(t);
if l=r then
begin
tree[t].m:=y;
tree[t].s:=y;
exit(t);
end
else begin
m:=(l+r) shr ;
q:=t;
if x<=m then
begin
tree[q].r:=tree[last].r;
last:=tree[last].l;
tree[q].l:=change(last,l,m,x,y);
end
else begin
tree[q].l:=tree[last].l;
last:=tree[last].r;
tree[q].r:=change(last,m+,r,x,y);
end;
update(q);
exit(q);
end;
end; function getmax(now,l,r,x,y:longint):longint;
var m,q:longint;
begin
if (x<=l) and (y>=r) then exit(tree[now].m)
else begin
m:=(l+r) shr ;
q:=;
if (x<=m) and (tree[now].l<>) then q:=getmax(tree[now].l,l,m,x,y);
if (y>m) and (tree[now].r<>) then q:=max(q,getmax(tree[now].r,m+,r,x,y));
exit(q);
end;
end; function getsum(now,l,r,x,y:longint):longint;
var m,q:longint;
begin
if (x<=l) and (y>=r) then exit(tree[now].s)
else begin
m:=(l+r) shr ;
q:=;
if (x<=m) and (tree[now].l<>) then q:=q+getsum(tree[now].l,l,m,x,y);
if (y>m) and (tree[now].r<>) then q:=q+getsum(tree[now].r,m+,r,x,y);
exit(q);
end;
end; function maxx(x,y:longint):longint;
var f1,f2,re:longint;
begin
maxx:=;
re:=b[x];
f1:=top[x];
f2:=top[y];
while f1<>f2 do
begin
if d[f1]>=d[f2] then
begin
maxx:=max(maxx,getmax(h[re],,n,c[f1],c[x]));
x:=fa[f1];
end
else begin
maxx:=max(maxx,getmax(h[re],,n,c[f2],c[y]));
y:=fa[f2];
end;
f1:=top[x];
f2:=top[y];
end;
if c[x]>c[y] then swap(x,y);
maxx:=max(maxx,getmax(h[re],,n,c[x],c[y]));
end; function ask(x,y:longint):longint;
var f1,f2,re:longint;
begin
ask:=;
re:=b[x];
f1:=top[x];
f2:=top[y];
while f1<>f2 do
begin
if d[f1]>=d[f2] then
begin
ask:=ask+getsum(h[re],,n,c[f1],c[x]);
x:=fa[f1];
end
else begin
ask:=ask+getsum(h[re],,n,c[f2],c[y]);
y:=fa[f2];
end;
f1:=top[x];
f2:=top[y];
end;
if c[x]>c[y] then swap(x,y);
ask:=ask+getsum(h[re],,n,c[x],c[y]);
end; begin
readln(n,q);
for i:= to n do
readln(w[i],b[i]);
for i:= to n- do
begin
readln(x,y);
add(x,y);
add(y,x);
end;
d[]:=;
dfs1();
top[]:=;
dfs2();
t:=;
for i:= to n do
h[b[i]]:=change(h[b[i]],,n,c[i],w[i]);
for i:= to q do
begin
read(ch);
if ch='C' then
begin
readln(ch,x,y);
if ch='C' then
begin
h[b[x]]:=change(h[b[x]],,n,c[x],);
b[x]:=y;
h[y]:=change(h[y],,n,c[x],w[x]);
end
else begin
w[x]:=y;
h[b[x]]:=change(h[b[x]],,n,c[x],y);
end;
end
else begin
readln(ch,x,y);
if ch='M' then
writeln(maxx(x,y))
else writeln(ask(x,y));
end;
end;
end.