若干个树， q个询问
操作1：求出某一棵树的直径；
操作2：合并两个没合并的树（若合并了则省略），合并的树中树的直径最小。

见画板：

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
const int N = 3e5 + 10, M = 2 * N, inf = 0x3f3f3f3f;

int n, m, q;
int h[N], e[M], ne[M], idx;
int p[N], pd[N];
int id, d;

void add(int a, int b)
{
    e[idx] = b, ne[idx] = h[a], h[a] = idx ++;
}

int find(int x)
{
    return x == p[x] ? p[x] : p[x] = find(p[x]);
}

void dfs(int u, int fa, int sum)
{
    if(sum > d)
    {
        d = sum;
        id = u;
    }
    for(int i = h[u]; ~i; i = ne[i])
    {
        int son = e[i];
        if(son == fa) continue;
        dfs(son, u, sum + 1);   
    }
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);

    memset(h, -1, sizeof h);
    for(int i = 0; i < N; i ++) p[i] = i, pd[i] = 0;

    cin >> n >> m >> q;
    for(int i = 0; i < m; i ++)
    {
        int a, b;
        cin >> a >> b;
        add(a, b), add(b, a);

        int pa = find(a), pb = find(b);
        if(pa != pb) p[pa] = pb;
    }
    vector<int> root;
    for(int i = 1; i <= n; i ++) 
    {
        if(i == p[i]) root.push_back(i);
    }
    for(int i = 0; i < root.size(); i ++)
    {
        id = root[i], d = 0;
        dfs(root[i], -1, 0);
        dfs(id, -1, 0);
        pd[root[i]] = d;
    }

    while(q --)
    {
        int c;
        cin >> c;
        if(c == 1)
        {
            int x;
            cin >> x;
            cout << pd[find(x)] << '\n';
        }
        else
        {
            int a, b;
            cin >> a >> b;
            int pa = find(a), pb = find(b);
            if(pa != pb)
            {
                int tmp = (pd[pa] + 1) / 2 + (pd[pb] + 1) / 2 + 1;
                p[pa] = pb;
                pd[pb] = max(tmp, max(pd[pa], pd[pb]));
            }
        }
    }
    return 0;
}