rt，本题我在用 lct 写的时候发现，如果将 makeroot 函数中的 rev[x] ^= 1 改成 rev[x] = 1 也能通过。
而且其他题目（P3366、P2417、P1501等）也行，这是数据太水了还是某种玄学因素？
(附：我的 lct 代码）

struct LCT{
	int ch[_N][2], fa[_N], sz[_N], rev[_N];
	#define ls ch[x][0]
	#define rs ch[x][1]
	#define get(x) (x == ch[find(fa[x])][1])
	#define isRoot(x) (ch[find(fa[x])][0] != x && ch[find(fa[x])][1] != x)

	void maintain(int x) { sz[0] = 0, sz[x] = sz[ls] + sz[rs] + 1; }
	void pushdown(int x) {
		if (rev[x]) {
			rev[ls] ^= 1, rev[rs] ^= 1;
			swap(ch[ls][0], ch[ls][1]);
			swap(ch[rs][0], ch[rs][1]);
			rev[x] = 0;
		}
	}
	void update(int x) {
		if (!isRoot(x)) update(find(fa[x]));
		pushdown(x);
	}
	void rotate(int x) {
		x = find(x);
		int y = find(fa[x]), z = find(fa[y]), tp = get(x);
		if (!isRoot(y)) ch[z][get(y)] = x;
		fa[x] = z;
		ch[y][tp] = ch[x][tp ^ 1];
		if (ch[x][tp ^ 1]) fa[ch[x][tp ^ 1]] = y;
		ch[x][tp ^ 1] = y;
		fa[y] = x;
		maintain(y);
		maintain(x);
	}
	void splay(int x) {
		x = find(x);
		update(x);
		while (!isRoot(x)) {
			int y = fa[x];
			if (!isRoot(y)) rotate(get(x) == get(y) ? y : x);
			rotate(x);
		}
	}
	void access(int x) {
		for (int p = 0; x; x = find(fa[p = x])) {
			splay(x);
			ch[x][1] = p;
			maintain(x);
		}
	}
	void makeRoot(int x) {
		x = find(x);
		access(x);
		splay(x);
		rev[x] = 1;
		swap(ls, rs);
	}
	int findRoot(int x) {
		x = find(x);
		access(x);
		splay(x);
		pushdown(x);
		while (ls) pushdown(x), x = ls;
		splay(x);
		return x;
	}
	void split(int x, int y) {
		makeRoot(x);
		access(y);
		splay(y);
	}
	void subtree(int x) {
		pushdown(x);
		if (ls) subtree(ls), merge(x, ls);
		if (rs) subtree(rs), merge(x, rs);
	}
	void link(int x, int y) {
		makeRoot(x);
		splay(x);
		fa[x] = y;
	}
}lct;```