bzoj 4551 [Tjoi2016&Heoi2016]树 树剖+线段树

题面

题目传送门

解法

并查集的方法感觉十分精妙，见dalao题解
树剖+线段树也比较简单吧

维护区间的答案，合并的时候根据深度判断大小

时间复杂度：\(O(q\ log^2\ n)\)

代码

#include <bits/stdc++.h>
#define N 100010
using namespace std;
template <typename node> void read(node &x) {
	x = 0; int f = 1; char c = getchar();
	while (!isdigit(c)) {if (c == '-') f = -1; c = getchar();}
	while (isdigit(c)) x = x * 10 + c - '0', c = getchar(); x *= f;
}
int n, q, cnt, Time;
int dfn[N], siz[N], top[N], d[N], f[N], son[N];
struct Edge {
	int next, num;
} e[N * 3];
struct SegmentTree {
	struct Node {
		int l, r, mx;
	} t[N * 4];
	void build(int k, int l, int r) {
		t[k] = (Node) {l, r, n + 1};
		if (l == r) return;
		int mid = (l + r) >> 1;
		build(k << 1, l, mid);
		build(k << 1 | 1, mid + 1, r);
	}
	void update(int k) {
		int x = t[k << 1].mx, y = t[k << 1 | 1].mx;
		t[k].mx = (d[x] > d[y]) ? x : y;
	}
	void modify(int k, int x, int num) {
		int l = t[k].l, r = t[k].r;
		if (l == r) {t[k].mx = num; return;}
		int mid = (l + r) >> 1;
		if (x <= mid) modify(k << 1, x, num);
			else modify(k << 1 | 1, x, num);
		update(k);
	}
	int query(int k, int L, int R) {
		int l = t[k].l, r = t[k].r;
		if (L <= l && r <= R) return t[k].mx;
		int mid = (l + r) >> 1;
		if (R <= mid) return query(k << 1, L, R);
		if (L > mid) return query(k << 1 | 1, L, R);
		int x = query(k << 1, L, mid), y = query(k << 1 | 1, mid + 1, R);
		return (d[x] > d[y]) ? x : y;
	}
} T;
void add(int x, int y) {
	e[++cnt] = (Edge) {e[x].next, y};
	e[x].next = cnt;
}
void dfs1(int x, int fa) {
	d[x] = d[fa] + 1, siz[x] = 1, f[x] = fa;
	for (int p = e[x].next; p; p = e[p].next) {
		int k = e[p].num;
		if (k == fa) continue; 
		dfs1(k, x); siz[x] += siz[k];
		if (siz[son[x]] < siz[k]) son[x] = k;
	}
}
void dfs2(int x, int tp) {
	top[x] = tp, dfn[x] = ++Time;
	if (!son[x]) return; dfs2(son[x], tp);
	for (int p = e[x].next; p; p = e[p].next) {
		int k = e[p].num;
		if (k == f[x] || k == son[x]) continue;
		dfs2(k, k);
	}
}
int Query(int x) {
	int fx = top[x];
	while (fx != 1) {
		int tmp = T.query(1, dfn[fx], dfn[x]);
		if (tmp != n + 1) return tmp;
		x = f[fx], fx = top[x];
	}
	return T.query(1, dfn[1], dfn[x]);
}
int main() {
	read(n), read(q); cnt = n;
	for (int i = 1; i < n; i++) {
		int x, y; read(x), read(y);
		add(x, y), add(y, x);
	}
	dfs1(1, 0); dfs2(1, 0); d[n + 1] = 0;
	T.build(1, 1, n); T.modify(1, dfn[1], 1);
	while (q--) {
		char c = getchar(); int x;
		while (!isalpha(c)) c = getchar();
		read(x);
		if (c == 'C') T.modify(1, dfn[x], x);
			else cout << Query(x) << "\n";
	}
	return 0;
}