$ \color{#0066ff}{ 题目描述 }$
给定一棵树，有m次操作。
1 x 把第x条边染成黑色
2 x 把第x条边染成白色
3 x y 查询x~y之间的黑边数，存在白边输出-1
\(\color{#0066ff}{输入格式}\)
第一行一个正整数N (1 ≤ N ≤ 100000)，节点总数
接下来N − 1行，每行两个整数a,b 表示一条边
接下来是一个正整数m(1 ≤ m ≤ 300000)，表示共有m次操作。
后面跟着m行，是操作
\(\color{#0066ff}{输出格式}\)
对于每一个询问，输出一行答案
\(\color{#0066ff}{输入样例}\)
3
1 2
2 3
7
3 1 2
3 1 3
3 2 3
2 2
3 1 2
3 1 3
3 2 3
6
1 5
6 4
2 3
3 5
5 6
6
3 3 4
2 5
3 2 6
3 1 2
2 3
3 3 1\(\color{#0066ff}{输出样例}\)
1
2
1
1
-1
-1
3
-1
3
2\(\color{#0066ff}{数据范围与提示}\)
none
\(\color{#0066ff}{题解}\)
只有两种颜色，维护两个LCT就行了
LCT上维护siz，询问时siz-1就是答案
无解就是不联通qwq
#include<bits/stdc++.h>
#define LL long long
LL in() {
char ch; LL x = 0, f = 1;
while(!isdigit(ch = getchar()))(ch == '-') && (f = -f);
for(x = ch ^ 48; isdigit(ch = getchar()); x = (x << 1) + (x << 3) + (ch ^ 48));
return x * f;
}
const int maxn = 1e5 + 10;
struct LCT {
protected:
struct node {
node *ch[2], *fa;
int siz, rev;
node(int siz = 0, int rev = 0): siz(siz), rev(rev) {
fa = ch[0] = ch[1] = NULL;
}
void trn() { std::swap(ch[0], ch[1]), rev ^= 1; }
void dwn() {
if(!rev) return;
if(ch[0]) ch[0]->trn();
if(ch[1]) ch[1]->trn();
rev = 0;
}
void upd() { siz = (ch[0]? ch[0]->siz : 0) + (ch[1]? ch[1]->siz : 0) + 1; }
bool isr() { return fa->ch[1] == this; }
bool ntr() { return fa && (fa->ch[1] == this || fa->ch[0] == this); }
}pool[maxn];
void rot(node *x) {
node *y = x->fa, *z = y->fa;
bool k = x->isr(); node *w = x->ch[!k];
if(y->ntr()) z->ch[y->isr()] = x;
(x->ch[!k] = y)->ch[k] = w;
(y->fa = x)->fa = z;
if(w) w->fa = y;
y->upd(), x->upd();
}
void splay(node *o) {
static node *st[maxn];
int top;
st[top = 1] = o;
while(st[top]->ntr()) st[top + 1] = st[top]->fa, top++;
while(top) st[top--]->dwn();
while(o->ntr()) {
if(o->fa->ntr()) rot(o->isr() ^ o->fa->isr()? o : o->fa);
rot(o);
}
}
void access(node *x) {
for(node *y = NULL; x; x = (y = x)->fa)
splay(x), x->ch[1] = y, x->upd();
}
void makeroot(node *x) { access(x), splay(x), x->trn(); }
node *findroot(node *x) {
access(x), splay(x);
while(x->dwn(), x->ch[0]) x = x->ch[0];
return x;
}
public:
void link(int l, int r) {
node *x = pool + l, *y = pool + r;
makeroot(x), x->fa = y;
}
void cut(int l, int r) {
node *x = pool + l, *y = pool + r;
makeroot(x), access(y), splay(y);
if(y->ch[0] == x) y->ch[0] = x->fa = NULL;
}
int query(int l, int r) {
node *x = pool + l, *y = pool + r;
if(findroot(x) != findroot(y)) return -1;
makeroot(x), access(y), splay(y);
return y->siz - 1;
}
}s[2];
std::pair<int, int> mp[maxn];
int col[maxn];
int main() {
int n = in();
for(int i = 1; i < n; i++) s[1].link(mp[i].first = in(), mp[i].second = in()), col[i] = 1;
int x, y, p;
for(int T = in(); T --> 0;) {
p = in();
if(p == 1) {
if(col[x = in()]) continue;
s[0].cut(mp[x].first, mp[x].second);
s[col[x] = 1].link(mp[x].first, mp[x].second);
}
if(p == 2) {
if(!col[x = in()]) continue;
s[1].cut(mp[x].first, mp[x].second);
s[col[x] = 0].link(mp[x].first, mp[x].second);
}
if(p == 3) x = in(), y = in(), printf("%d\n", s[1].query(x, y));
}
return 0;
}