题面
解法
挺恶心的题
考虑动态点分治，先建出点分树
然后每一个点开两个堆，分别为\(a,b\)
\(a_i\)表示点分树上\(i\)子树中所有节点在原树上和点分树中\(i\)父亲的距离，\(b_i\)表示点分树中\(i\)所有儿子的堆顶
再开一个堆\(ans\)，存每一个\(b_i\)最大和次大值的和
在修改的时候，分两种情况考虑
但是本质都是一样的，就是在点分树上不断爬，爬到根为止，然后对当前点和父亲的\(a,b\)进行删除和加入
细节比较多，需要注意
重点：传入一个结构体参数的时候一定要加&，否则会T成皮皮
时间复杂度：\(O(q\ log^2\ n)\)
代码
#include <bits/stdc++.h>
#define inf 1 << 30
#define N 100010
using namespace std;
template <typename node> void chkmax(node &x, node y) {x = max(x, y);}
template <typename node> void chkmin(node &x, node y) {x = min(x, y);}
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;
}
struct Heap {
priority_queue <int> ret, del;
void push(int x) {ret.push(x);}
void erase(int x) {del.push(x);}
int siz() {return ret.size() - del.size();}
void Pop() {
while (!ret.empty() && !del.empty() && ret.top() == del.top())
ret.pop(), del.pop();
if (ret.size()) ret.pop();
}
int top() {
if (!siz()) return 0;
while (!ret.empty() && !del.empty() && ret.top() == del.top())
ret.pop(), del.pop();
return ret.top();
}
int setop() {
if (siz() < 2) return 0;
int x = top(); Pop();
int y = top(); push(x);
return y;
}
} a[N], b[N], ans;
struct Edge {
int next, num;
} e[N * 3];
int n, tot, cnt, rt, now, c[N], d[N], f[N], p[N], dep[N], siz[N], vis[N], ff[N][20];
vector <int> E[N];
void add(int x, int y) {
e[++cnt] = (Edge) {e[x].next, y};
e[x].next = cnt;
}
void getr(int x, int fa) {
f[x] = 0, siz[x] = 1;
for (int i = 0; i < E[x].size(); i++) {
int k = E[x][i];
if (k == fa || vis[k]) continue;
getr(k, x); chkmax(f[x], siz[k]);
siz[x] += siz[k];
}
chkmax(f[x], now - siz[x]);
if (f[x] < f[rt]) rt = x;
}
void work(int x, int fa) {
vis[x] = 1, p[x] = fa;
if (fa) add(fa, x);
for (int i = 0; i < E[x].size(); i++) {
int k = E[x][i];
if (vis[k]) continue;
f[rt = 0] = inf, now = siz[k];
getr(k, x); work(rt, x);
}
}
void dfs(int x, int fa) {
d[x] = d[fa] + 1;
for (int i = 1; i <= 18; i++)
ff[x][i] = ff[ff[x][i - 1]][i - 1];
for (int i = 0; i < E[x].size(); i++) {
int k = E[x][i];
if (k == fa) continue; ff[k][0] = x;
dfs(k, x);
}
}
int lca(int x, int y) {
if (d[x] < d[y]) swap(x, y);
for (int i = 18; i >= 0; i--)
if (d[ff[x][i]] >= d[y]) x = ff[x][i];
if (x == y) return x;
for (int i = 18; i >= 0; i--)
if (ff[x][i] != ff[y][i]) x = ff[x][i], y = ff[y][i];
return ff[x][0];
}
int dis(int x, int y) {
int z = lca(x, y);
return d[x] + d[y] - 2 * d[z];
}
void update(int x, int y, int z) {
a[z].push(dis(x, y));
for (int p = e[x].next; p; p = e[p].next)
update(e[p].num, y, z);
}
void getd(int x) {
dep[x] = dep[p[x]] + 1;
for (int p = e[x].next; p; p = e[p].next)
getd(e[p].num);
}
void init() {
f[rt = 0] = inf;
dfs(1, 0); now = n;
getr(1, 0); work(rt, 0);
for (int i = 1; i <= n; i++)
if (p[i]) update(i, p[i], i);
for (int i = 1; i <= n; i++) {
b[i].push(0);
for (int p = e[i].next; p; p = e[p].next) {
int k = e[p].num;
if (a[k].siz()) b[i].push(a[k].top());
}
}
for (int i = 1; i <= n; i++)
ans.push(b[i].top() + b[i].setop());
}
void Insert(Heap &a) {
if (a.siz() >= 2) {
int t = a.top() + a.setop();
ans.push(t);
}
}
void Erase(Heap &a) {
if (a.siz() >= 2) {
int t = a.top() + a.setop();
ans.erase(t);
}
}
void modify(int x) {
if (c[x] == 1) {
Erase(b[x]), b[x].push(0);
Insert(b[x]);
for (int y = x; p[y]; y = p[y]) {
Erase(b[p[y]]);
if (a[y].siz()) b[p[y]].erase(a[y].top());
a[y].push(dis(p[y], x));
if (a[y].siz()) b[p[y]].push(a[y].top());
Insert(b[p[y]]);
}
tot++;
} else {
Erase(b[x]); b[x].erase(0);
Insert(b[x]);
for (int y = x; p[y]; y = p[y]) {
Erase(b[p[y]]);
if (a[y].siz()) b[p[y]].erase(a[y].top());
a[y].erase(dis(p[y], x));
if (a[y].siz()) b[p[y]].push(a[y].top());
Insert(b[p[y]]);
}
tot--;
}
c[x] ^= 1;
}
int main() {
read(n); cnt = tot = n;
for (int i = 1; i < n; i++) {
int x, y; read(x), read(y);
E[x].push_back(y), E[y].push_back(x);
}
init(); int q; read(q);
while (q--) {
char c = getchar();
while (!isalpha(c)) c = getchar();
if (c == 'C') {
int x; read(x);
modify(x);
} else
if (tot < 2) cout << tot - 1 << "\n";
else cout << ans.top() << "\n";
}
return 0;
}