题解 连通块

传送门

将询问离线，倒序枚举将删边转化为加边
并查集维护连通块内直径即可

Code:

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 200010
#define ll long long
#define fir first
#define sec second
#define make make_pair
//#define int long long

char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n, m;
int head[N], size;
int op[N], x[N];
bool del[N];
pair<int, int> e2[N];
struct edge{int to, next;}e[N<<1];
inline void add(int s, int t) {e[++size].to=t; e[size].next=head[s]; head[s]=size;}

namespace task1{
	int fa[24][N], dep[N], lg[N], dsu[N], s1[N], s2[N], sta[N], top;
	inline int find(int p) {return dsu[p]==p?p:dsu[p]=find(dsu[p]);}
	void dfs(int u, int pa) {
		for (int i=1; i<=23; ++i)
			if (dep[u]>=1<<i) fa[i][u]=fa[i-1][fa[i-1][u]];
			else break;
		for (int i=head[u],v; ~i; i=e[i].next) {
			v = e[i].to;
			if (v==pa) continue;
			dep[v]=dep[u]+1; fa[0][v]=u;
			dfs(v, u);
		}
	}
	int lca(int a, int b) {
		if (dep[a]<dep[b]) swap(a, b);
		while (dep[a]>dep[b]) a=fa[lg[dep[a]-dep[b]]-1][a];
		if (a==b) return a;
		for (int i=lg[dep[a]]-1; ~i; --i)
			if (fa[i][a]!=fa[i][b])
				a=fa[i][a], b=fa[i][b];
		return fa[0][a];
	}
	inline int dis(int a, int b) {return dep[a]+dep[b]-2*dep[lca(a, b)];}
	void uni(int s, int t) {
		s=find(s); t=find(t);
		if (s==t) return ;
		dsu[s]=t;
		int d11=dis(s1[s], s1[t]), d12=dis(s1[s], s2[t]), d21=dis(s2[s], s1[t]), d22=dis(s2[s], s2[t]), ds=dis(s1[s], s2[s]), dt=dis(s1[t], s2[t]), maxn=0;
		maxn=max(maxn, d11); maxn=max(maxn, d12); maxn=max(maxn, d21); maxn=max(maxn, d22); maxn=max(maxn, ds); maxn=max(maxn, dt);
		int t1=0, t2=0;
		if (d11==maxn) t1=s1[s], t2=s1[t];
		else if (d12==maxn) t1=s1[s], t2=s2[t];
		else if (d21==maxn) t1=s2[s], t2=s1[t];
		else if (d22==maxn) t1=s2[s], t2=s2[t];
		else if (ds==maxn) t1=s1[s], t2=s2[s];
		else if (dt==maxn) t1=s1[t], t2=s2[t];
		else puts("error");
		s1[t]=t1, s2[t]=t2;
	}
	void solve() {
		for (int i=1; i<=n; ++i) lg[i]=lg[i-1]+(1<<lg[i-1]==i);
		for (int i=1; i<=n; ++i) dsu[i]=s1[i]=s2[i]=i;
		dep[1]=1; dfs(1, 0);
		for (int i=1; i<n; ++i) if (!del[i]) uni(e2[i].fir, e2[i].sec);
		for (int i=m; i; --i) {
			if (op[i]==1) uni(e2[x[i]].fir, e2[x[i]].sec);
			else {
				int ans=0, f=find(x[i]);
				ans=max(ans, dis(x[i], s1[f]));
				ans=max(ans, dis(x[i], s2[f]));
				sta[++top]=ans;
			}
		}
		while (top) printf("%d\n", sta[top--]);
		exit(0);
	}
}

signed main()
{
	freopen("block.in", "r", stdin);
	freopen("block.out", "w", stdout);

	n=read(); m=read();
	memset(head, -1, sizeof(head));
	for (int i=1,u,v; i<n; ++i) {
		u=read(); v=read();
		add(u, v); add(v, u);
		e2[i]=make(u, v);
	}
	for (int i=1; i<=m; ++i) {
		op[i]=read(); x[i]=read();
		if (op[i]==1) del[x[i]]=1;
	}
	task1::solve();

	return 0;
}