原题传送门
对于两个连通块连起来，肯定是在两个直径中点之间连一条边使得新的直径最短
并查集维护连通性，同时记录连通块的直径大小
令两个连通块的直径分别为 l e n s 1 > l e n s 2 len_{s1}>len_{s2} lens1>lens2
则新的直径是 m a x ( l e n s 1 , l e n s 1 + 1 2 + l e n s 2 + 1 2 + 1 ) max(len_{s1},\frac{len_{s1}+1}{2}+\frac{len_{s2}+1}{2}+1) max(lens1,2lens1+1+2lens2+1+1)
除法是上取整的意思
Code：
#include <bits/stdc++.h>
#define maxn 300010
using namespace std;
struct Edge{
int to, next;
}edge[maxn << 1];
int num, head[maxn], rt, Max;
int n, m, q, f[maxn], len[maxn];
inline int read(){
int s = 0, w = 1;
char c = getchar();
for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
return s * w;
}
void addedge(int x, int y){
edge[++num] = (Edge){
y, head[x]}, head[x] = num; }
int getfa(int k){
return f[k] == k ? k : f[k] = getfa(f[k]); }
void dfs(int u, int pre, int s){
if (s > Max) Max = s, rt = u;
for (int i = head[u]; i; i = edge[i].next){
int v = edge[i].to;
if (v != pre) dfs(v, u, s + 1);
}
}
void merge(int x, int y){
int s1 = getfa(x), s2 = getfa(y);
if (len[s1] < len[s2]) swap(s1, s2);
f[s2] = s1, len[s1] = max(len[s1], (len[s1] + 1) / 2 + (len[s2] + 1) / 2 + 1);
}
int main(){
// freopen("1.txt", "r", stdin);
// freopen("21.out", "w", stdout);
n = read(), m = read(), q = read();
for (int i = 1; i <= n; ++i) f[i] = i;
while (m--){
int x = read(), y = read(), s1 = getfa(x), s2 = getfa(y);
addedge(x, y), addedge(y, x);
f[s1] = s2;
}
for (int i = 1; i <= n; ++i){
int x = getfa(i);
if (!len[x]){
rt = 0, Max = 0;
dfs(i, 0, 0);
Max = 0;
dfs(rt, 0, 0);
len[x] = Max;
}
}
while (q--){
int opt = read();
if (opt == 1) printf("%d\n", len[getfa(read())]); else{
int x = read(), y = read();
if (getfa(x) != getfa(y)) merge(x, y);
}
}
return 0;
}