题面
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;
}