SDOI2017 树点涂色（LCT+线段树）

题目链接

题目大意

给定一棵每个点颜色不同的树，支持三种操作：
1.把点x到根节点的路径上的所有的点染上一种没有用过的新颜色；
2.求x到y有多少种颜色；
3.求x的子树中的点使该点到根的不同颜色数最大。
$n,m\le 10^5$

题解

如果我们能够动态维护某个点到根的不同颜色数，就做完了。考虑1操作实际上是LCT的access操作，于是我们可以用LCT来维护。
access时每次会断开一条边并连上一条边，也就是让断开的儿子所在子树颜色数+1，连上的儿子所在子树颜色数-1.
因此直接用线段树维护dfs序即可。复杂度 $O(nlog^2n)$。

#include <bits/stdc++.h>
using namespace std;
const int MAXR = 10000000;
char _READ_[MAXR], _PRINT_[MAXR];
int _READ_POS_, _PRINT_POS_, _READ_LEN_;
inline char readc() {
#ifndef ONLINE_JUDGE
	return getchar();
#endif
	if (!_READ_POS_) _READ_LEN_ = fread(_READ_, 1, MAXR, stdin);
	char c = _READ_[_READ_POS_++];
	if (_READ_POS_ == MAXR) _READ_POS_ = 0;
	if (_READ_POS_ > _READ_LEN_) return 0;
	return c;
}
template<typename T> inline void read(T &x) {
	x = 0; register int flag = 1, c;
	while (((c = readc()) < '0' || c > '9') && c != '-');
	if (c == '-') flag = -1; else x = c - '0';
	while ((c = readc()) >= '0' && c <= '9') x = x * 10 + c - '0';
	x *= flag;
}
template<typename T1, typename ...T2> inline void read(T1 &a, T2&... x) {
	read(a), read(x...);
}
inline int reads(char *s) {
	register int len = 0, c;
	while (isspace(c = readc()) || !c);
	s[len++] = c;
	while (!isspace(c = readc()) && c) s[len++] = c;
	s[len] = 0;
	return len;
}
inline void ioflush() { fwrite(_PRINT_, 1, _PRINT_POS_, stdout), _PRINT_POS_ = 0; fflush(stdout); }
inline void printc(char c) {
	_PRINT_[_PRINT_POS_++] = c;
	if (_PRINT_POS_ == MAXR) ioflush();
}
inline void prints(char *s) {
	for (int i = 0; s[i]; i++) printc(s[i]);
}
template<typename T> inline void print(T x, char c = '\n') {
	if (x < 0) printc('-'), x = -x;
	if (x) {
		static char sta[20];
		register int tp = 0;
		for (; x; x /= 10) sta[tp++] = x % 10 + '0';
		while (tp > 0) printc(sta[--tp]);
	} else printc('0');
	printc(c);
}
template<typename T1, typename ...T2> inline void print(T1 x, T2... y) {
	print(x, ' '), print(y...);
}
typedef long long ll;

const int MAXN = 100005, MAXT = 262150;
struct Edge { int to, next; } edge[MAXT];
int head[MAXN], son[MAXN][2], beg[MAXN], ed[MAXN], dep[MAXN];
int mx[MAXT], tag[MAXT], ppar[20][MAXN], par[MAXN], n, m, tot, tn;
void addedge(int u, int v) {
	edge[++tot] = (Edge) { v, head[u] };
	head[u] = tot;
}
void dfs(int u, int fa) {
	beg[u] = tot++, par[u] = fa, dep[u] = dep[fa] + 1;
	for (int i = head[u]; i; i = edge[i].next) {
		int v = edge[i].to;
		if (v != fa) dfs(v, u);
	}
	ed[u] = tot;
}
void pushdown(int k) {
	if (!tag[k]) return;
	int ls = k * 2 + 1, rs = ls + 1;
	tag[ls] += tag[k], tag[rs] += tag[k];
	mx[ls] += tag[k], mx[rs] += tag[k];
	tag[k] = 0;
}
void update(int a, int b, int x, int l = 0, int r = tn, int k = 0) {
	if (a >= r || b <= l) return;
	if (a <= l && b >= r) { tag[k] += x, mx[k] += x; return; }
	pushdown(k);
	int mid = (l + r) >> 1, ls = k * 2 + 1, rs = ls + 1;
	update(a, b, x, l, mid, ls);
	update(a, b, x, mid, r, rs);
	mx[k] = max(mx[ls], mx[rs]);
}
int query(int a, int b, int l = 0, int r = tn, int k = 0) {
	if (a >= r || b <= l) return 0;
	if (a <= l && b >= r) return mx[k];
	pushdown(k);
	int mid = (l + r) >> 1, ls = k * 2 + 1;
	return max(query(a, b, l, mid, ls), query(a, b, mid, r, ls + 1));
}
bool isroot(int x) { int y = par[x]; return !y || (son[y][0] != x && son[y][1] != x); }
void rotate(int x) {
	int y = par[x], z = par[y], w = son[y][0] == x, t = son[x][w];
	if (!isroot(y)) son[z][son[z][1] == y] = x;
	par[x] = z, par[y] = x, son[y][!w] = t;
	if (t) par[t] = y; son[x][w] = y;
}
void splay(int x) {
	while (!isroot(x)) {
		int y = par[x], z = par[y];
		if (!isroot(y)) {
			if ((son[z][1] == y) ^ (son[y][1] == x)) rotate(x);
			else rotate(y);
		} rotate(x);
	}
}
void access(int x) {
	int y = 0;
	for (; x; x = par[x]) {
		splay(x);
		int &t = son[x][1];
		if (t) {
			int z = t;
			while (son[z][0]) z = son[z][0];
			update(beg[z], ed[z], 1);
		}
		if (y) {
			int z = y;
			while (son[z][0]) z = son[z][0];
			update(beg[z], ed[z], -1);
		}
		t = y, y = x;
	}
}
int get_lca(int x, int y) {
	if (dep[x] > dep[y]) swap(x, y);
	for (int i = 19; i >= 0; i--) if (dep[ppar[i][y]] >= dep[x]) y = ppar[i][y];
	if (x == y) return x;
	for (int i = 19; i >= 0; i--)
		if (ppar[i][x] != ppar[i][y]) x = ppar[i][x], y = ppar[i][y];
	return ppar[0][x];
}
int main() {
	read(n, m);
	for (int i = 1; i < n; i++) {
		int u, v; read(u, v);
		addedge(u, v), addedge(v, u);
	}
	dfs(1, tot = 0);
	memcpy(ppar[0], par, sizeof(ppar[0]));
	for (int i = 1; i < 20; i++)
	for (int j = 1; j <= n; j++)
		ppar[i][j] = ppar[i - 1][ppar[i - 1][j]];
	for (tn = 1; tn < n; tn <<= 1);
	for (int i = 1; i <= n; i++) update(beg[i], ed[i], 1);
	while (m--) {
		int opt, x, y;
		read(opt, x);
		if (opt == 1) access(x);
		else if (opt == 2) {
			read(y);
			int l = get_lca(x, y); splay(l);
			int res = query(beg[x], beg[x] + 1) + query(beg[y], beg[y] + 1);
			if (par[l]) res -= query(beg[par[l]], beg[par[l]] + 1) * 2;
			print(res - 1);
		} else print(query(beg[x], ed[x]));
	}
	ioflush();
	return 0;
}