参考了皎月半洒花的博客

看到树想到树剖，由于要取距自己到根离自己最近的标记点，刚开始想到线段树里存节点深度，查询时返回最大值。但是这样的话只能得到节点深度，无法得知节点编号，就想倍增乱搞一下，求出标记点，复杂度\(O(\log ^ {3}\;N)\)

虽然可以过但是实现有点复杂，就看了一下上面的博客

真的很强，由于树剖dfs时一条链上的编号是连续的，在此链中且深度越大线段树编号越大，所以我们可以在线段树里存当前节点的线段树编号，也达到了维护深度最大值的效果

答案就是ori [ (一条链中) MAX index] (ori为线段树编号回找树原始编号的数组)

复杂度\(O(\log ^ {2}\;N)\)

一直都是把树剖当板子用的，现在发现结合性质还有更多用处，我还要加油啊

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
int RD(){int out = 0,flag = 1;char c = getchar();while(c < '0' || c >'9'){if(c == '-')flag = -1;c = getchar();}while(c >= '0' && c <= '9'){out = out * 10 + c - '0';c = getchar();}return flag * out;}
const int maxn = 1000019,INF = 1e9;
int num,na,nume,cnt;
int head[maxn];
struct Node{int v,nxt;}E[maxn * 2];
void add(int u,int v){E[++nume].nxt = head[u];E[nume].v = v;head[u] = nume;}
int size[maxn],wson[maxn],dep[maxn],fa[maxn],top[maxn],pos[maxn],ori[maxn];
int v[maxn];
void dfs1(int id,int F){size[id] = 1;for(int i = head[id];i;i = E[i].nxt){int v = E[i].v;if(v == F)continue;dep[v] = dep[id] + 1;fa[v] = id;dfs1(v,id);size[id] += size[v];if(size[v] > size[wson[id]])wson[id] = v;}}
void dfs2(int id,int TP){top[id] = TP;pos[id] = ++cnt;ori[cnt] = id;if(!wson[id])return ;dfs2(wson[id],TP);for(int i = head[id];i;i = E[i].nxt){int v = E[i].v;if(v == fa[id] || v == wson[id])continue;dfs2(v,v);}}
#define lid (id << 1)
#define rid (id << 1) | 1
struct sag_tree{int l,r,max;}tree[maxn << 2];
void build(int id,int l,int r){tree[id].l = l;tree[id].r = r;if(l == r){tree[id].max = 0;return ;}int mid = (l + r) >> 1;build(lid,l,mid);build(rid,mid + 1,r);tree[id].max = max(tree[lid].max,tree[rid].max);}
void update(int id,int val, int l,int r){if(tree[id].l == l && tree[id].r == r){tree[id].max = l;return ;}int mid = (tree[id].l + tree[id].r) >> 1;if(mid < l)update(rid,val,l,r);else if(mid >= r)update(lid,val,l,r);else update(lid,val,l,mid),update(rid,val,mid + 1,r);tree[id].max = max(tree[lid].max,tree[rid].max);}
int query(int id,int l,int r){if(tree[id].l == l && tree[id].r == r)return tree[id].max;int mid = (tree[id].l + tree[id].r) >> 1;if(mid < l)return query(rid,l,r);else if(mid >= r)return query(lid,l,r);else return max(query(lid,l,mid),query(rid,mid + 1,r));}
void Qmax(int x, int y){int ans = 0;while(top[x] != top[y]){if(dep[top[x]] < dep[top[y]])swap(x, y);ans = query(1, pos[top[x]], pos[x]);if(ans){printf("%d\n", ori[ans]);return ;}x = fa[top[x]];}if(dep[x] > dep[y])swap(x, y);ans = query(1, pos[x], pos[y]);printf("%d\n", ori[ans]);}
int main(){num = RD();na = RD();for(int i = 1;i <= num - 1;i++){int u = RD(),v = RD();add(u,v),add(v,u);}dep[1] = 1;dfs1(1,-1);dfs2(1,1);build(1,1,num);update(1,pos[1],pos[1],pos[1]);for(int i = 1;i <= na;i++){char cmd;cin>>cmd;if(cmd == 'C'){int x = RD();update(1,pos[x],pos[x],pos[x]);}else{int x = RD();Qmax(x,1);}}return 0;}