其实这道题和以前在poj上做过的将树映射到树状数组的题目很像
首先不难想到，将一条边从土路修成公路，只对以这条边连接的孩子结点为根的子树有影响；
于是和之前那道poj的题目很像，后序遍历树，对每个节点重标号，每个点初始值就是深度
下面的问题就变成了：
土路修成公路---->区间修改
查询从点1到某个点所经过的土路数----->单点求值；
这种问题我们其实可以用树状数组来做；
a[i]表示原数组的值；
令c[i]=a[i]-a[i-1],特殊的c[1]=a[1];
然后我们对c数组做树状数组
区间修改（假设是[l,r]都+1）就是c[l]+1,c[r+1]-1;
单点求值就是求signma(c[1~i])
当然后来知道其实用dfs序更简单
type node=record
point,next:longint;
end; var fa,a,c,p,r,h,d:array[..] of longint;
edge:array[..] of node;
len,n,m,x,y,t,i:longint;
ch:char; procedure add(x,y:longint);
begin
inc(len);
edge[len].point:=y;
edge[len].next:=p[x];
p[x]:=len;
end; function lowbit(x:longint):longint;
begin
exit(x and (-x));
end; procedure work(x,f:longint);
begin
while x<=n do
begin
inc(a[x],f);
x:=x+lowbit(x);
end;
end; function ask(x:longint):longint;
begin
ask:=;
while x> do
begin
ask:=ask+a[x];
x:=x-lowbit(x);
end;
end; procedure dfs(x,d:longint);
var i,y,tmp:longint;
begin
i:=p[x];
c[x]:=d;
tmp:=n+;
while i<>- do
begin
y:=edge[i].point;
if (c[y]=) and (y<>) then
begin
fa[y]:=x;
dfs(y,d+);
if tmp>h[y] then tmp:=h[y];
end;
i:=edge[i].next;
end;
inc(t);
r[x]:=t;
if tmp=n+ then h[x]:=r[x]
else h[x]:=tmp;
end; begin
len:=-;
fillchar(p,sizeof(p),);
readln(n);
for i:= to n- do
begin
readln(x,y);
add(x,y);
add(y,x);
end;
t:=;
dfs(,);
for i:= to n do
d[r[i]]:=c[i];
for i:= to n do
work(i,d[i]-d[i-]);
work(,d[]);
readln(m);
for i:= to n+m- do
begin
read(ch);
if ch='W' then
begin
readln(x);
writeln(ask(r[x]));
end
else begin
readln(x,y);
if fa[x]=y then t:=x else t:=y;
work(h[t],-);
work(r[t]+,);
end;
end;
end.