[题解]LuoGu4092：[HEOI2016/TJOI2016]树

原题传送门
又见树剖裸题！
想想树剖的性质，询问一个点，往上跳，跳的一段区间一定是连续的，只要在这段区间中找出最大值就是最近的点
所以只要开一个维护最大值的线段树就行啦
另外需要注意根节点初始有标记

Code：

#include <bits/stdc++.h>
#define maxn 100010
#define ls rt << 1
#define rs rt << 1 | 1
using namespace std;
struct Edge{
	int to, next;
}edge[maxn << 1];
struct Seg{
	int l, r, sum;
}seg[maxn << 2];
int head[maxn], num, d[maxn], fa[maxn], size[maxn], son[maxn], id[maxn], cnt, top[maxn], rk[maxn], n, m;

inline int read(){
	int s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
}

void add_edge(int x, int y){ edge[++num].to = y; edge[num].next = head[x]; head[x] = num; }

void dfs(int u){
	size[u] = 1, son[u] = -1;
	for (int i = head[u]; i; i = edge[i].next){
		int v = edge[i].to;
		if (v != fa[u]){
			fa[v] = u;
			dfs(v);
			size[u] += size[v];
			if (son[u] == -1 || son[u] != -1 && size[son[u]] < size[v]) son[u] = v;
		}
	}
}

void dfs(int u, int x){
	id[u] = ++cnt, top[u] = x, rk[cnt] = u;
	if (son[u] == -1) return;
	dfs(son[u], x);
	for (int i = head[u]; i; i = edge[i].next){
		int v = edge[i].to;
		if (v != fa[u] && v != son[u]) dfs(v, v);
	}
}

void pushup(int rt){ seg[rt].sum = max(seg[ls].sum, seg[rs].sum); }

void build(int rt, int l, int r){
	seg[rt].l = l, seg[rt].r = r;
	if (l == r){
		if (l == 1) seg[rt].sum = 1;
		return;
	}
	int mid = (l + r) >> 1;
	build(ls, l, mid); build(rs, mid + 1, r);
	pushup(rt);
}

void update(int rt, int x){
	if (seg[rt].l == seg[rt].r){
		seg[rt].sum = x; return;
	}
	if (seg[ls].r >= x) update(ls, x); else update(rs, x);
	pushup(rt);
}

int query(int rt, int l, int r){
	if (seg[rt].l > r || seg[rt].r < l) return 0;
	if (seg[rt].l >= l && seg[rt].r <= r) return seg[rt].sum;
	return max(query(ls, l, r), query(rs, l, r));
}

int main(){
	n = read(), m = read();
	for (int i = 1; i < n; ++i){
		int x = read(), y = read();
		add_edge(x, y); add_edge(y, x);
	}
	dfs(1); dfs(1, 1); build(1, 1, n);
	while (m--){
		char c = getchar(); for (; c != 'Q' && c != 'C'; c = getchar()); int x = read();
		if (c == 'C') update(1, id[x]); else{
			int ans = 0;
			while (top[x] != 1){
				int s = query(1, id[top[x]], id[x]);
				if (s) ans = s;
				if (ans) break;
				x = fa[top[x]];
			}
			if (!ans) ans = query(1, id[top[x]], id[x]);
			printf("%d\n", rk[ans]);
		}
	}
	return 0;
}

当然这道题也可以用暴力线段树解决
Code：

#include <bits/stdc++.h>
#define maxn 100010
#define ls rt << 1
#define rs rt << 1 | 1
using namespace std;
struct Edge{
	int to, next;
}edge[maxn << 1];
struct Seg{
	int l, r, d, node;
}seg[maxn << 2];
int num, head[maxn], d[maxn], size[maxn], id[maxn], Index, n, m;

inline int read(){
	int s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
}

void addedge(int x, int y){ edge[++num] = (Edge){y, head[x]}, head[x] = num; }

void dfs(int u, int pre){
	d[u] = d[pre] + 1, size[u] = 1, id[u] = ++Index;
	for (int i = head[u]; i; i = edge[i].next){
		int v = edge[i].to;
		if (v != pre) dfs(v, u), size[u] += size[v];
	}
}

void pushdown(int rt){
	if (seg[ls].d < seg[rt].d) seg[ls].d = seg[rt].d, seg[ls].node = seg[rt].node;
	if (seg[rs].d < seg[rt].d) seg[rs].d = seg[rt].d, seg[rs].node = seg[rt].node;
}

void build(int rt, int l, int r){
	seg[rt].l = l, seg[rt].r = r;
	seg[rt].d = 1, seg[rt].node = 1;
	if (l == r) return;
	int mid = (l + r) >> 1;
	build(ls, l, mid), build(rs, mid + 1, r);
}

void update(int rt, int l, int r, int x){
	if (seg[rt].l > r || seg[rt].r < l) return;
	if (seg[rt].l >= l && seg[rt].r <= r){
		if (d[x] > seg[rt].d) seg[rt].d = d[x], seg[rt].node = x;
		return;
	}
	pushdown(rt);
	update(ls, l, r, x), update(rs, l, r, x);
}

int query(int rt, int pos){
	if (seg[rt].l == seg[rt].r) return seg[rt].node;
	pushdown(rt);
	if (pos <= seg[ls].r) return query(ls, pos); else return query(rs, pos);
}

int main(){
	n = read(), m = read();
	for (int i = 1; i < n; ++i){
		int x = read(), y = read();
		addedge(x, y), addedge(y, x);
	}
	dfs(1, 0);
	build(1, 1, n);
	while (m--){
		char opt = getchar();
		for (; opt != 'C' && opt != 'Q'; opt = getchar());
		int x = read();
		if (opt == 'C') update(1, id[x], id[x] + size[x] - 1, x); else printf("%d\n", query(1, id[x]));
	}
	return 0;
}