树的直径用两遍dfs扫一下就可以搞出来，预处理每一块的直径是2*n
考虑用并查集维护连通性，把答案保存在每个集合的根节点那儿即可
两棵树的直径合并起来后最小值如下：
max(ans[_find(x)],ans[_find(y)],(ans[_find(x)]+1)/2+(ans[_find(y)]+1)/2+1);
注意合并两棵树的时候判断一下是不是已经在一个集合里面了，否则会累加很多不存在的答案
代码：
#include <bits/stdc++.h> #define int long long #define sc(a) scanf("%lld",&a) #define scc(a,b) scanf("%lld %lld",&a,&b) #define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c) #define scs(a) scanf("%s",a) #define schar(a) scanf("%c",&a) #define pr(a) printf("%lld",a) #define fo(i,a,b) for(int i=a;i<b;++i) #define re(i,a,b) for(int i=a;i<=b;++i) #define rfo(i,a,b) for(int i=a;i>b;--i) #define rre(i,a,b) for(int i=a;i>=b;--i) #define prn() printf("\n") #define prs() printf(" ") #define mkp make_pair #define pii pair<int,int> #define pub(a) push_back(a) #define pob() pop_back() #define puf(a) push_front(a) #define pof() pop_front() #define fst first #define snd second #define frt front() #define bak back() #define mem0(a) memset(a,0,sizeof(a)) #define memmx(a) memset(a,0x3f3f,sizeof(a)) #define memmn(a) memset(a,-0x3f3f,sizeof(a)) #define debug #define db double #define yyes cout<<"YES"<<endl; #define nno cout<<"NO"<<endl; #define all(i,a) for(auto i=a.begin();i!=a.end();++i) using namespace std; typedef vector<int> vei; typedef vector<pii> vep; typedef map<int,int> mpii; typedef map<char,int> mpci; typedef map<string,int> mpsi; typedef deque<int> deqi; typedef deque<char> deqc; typedef priority_queue<int> mxpq; typedef priority_queue<int,vector<int>,greater<int> > mnpq; typedef priority_queue<pii> mxpqii; typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii; const int maxn=500005; const int inf=0x3f3f3f3f3f3f3f3f; const int MOD=100000007; const db eps=1e-10; const db pi=3.1415926535; int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;} int lowbit(int x){return x&-x;} int max(int a,int b){return a>b?a:b;} int min(int a,int b){return a<b?a:b;} int mmax(int a,int b,int c){return max(a,max(b,c));} int mmin(int a,int b,int c){return min(a,min(b,c));} void mod(int &a){a+=MOD;a%=MOD;} bool chk(int now){} int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;} int ll(int p){return p<<1;} int rr(int p){return p<<1|1;} int mm(int l,int r){return (l+r)/2;} int lg(int x){if(x==0) return 1;return (int)log2(x)+1;} bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;} bool bigeql(db a,db b){if(a>b||fabs(a-b)<=eps)return true;return false;} bool eql(db a,db b){if(fabs(a-b)<eps) return 1;return 0;} db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));} bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;} inline int read(){ char ch=getchar();int s=0,w=1; while(ch<48||ch>57){if(ch=='-')w=-1;ch=getchar();} while(ch>=48&&ch<=57){s=(s<<1)+(s<<3)+ch-48;ch=getchar();} return s*w; } inline void write(int x){ if(x<0)putchar('-'),x=-x; if(x>9)write(x/10); putchar(x%10+48); } int gcd(int a, int b){ if(a==0) return b; if(b==0) return a; if(!(a&1)&&!(b&1)) return gcd(a>>1,b>>1)<<1; else if(!(b&1)) return gcd(a,b>>1); else if(!(a&1)) return gcd(a>>1,b); else return gcd(abs(a-b),min(a,b)); } int lcm(int x,int y){return x*y/gcd(x,y);} int n,m,q,op,x,y,rt=0,mx=-inf; int f[maxn],ans[maxn]; bool vis[maxn]; vei e[maxn]; int _find(int x){ if(x!=f[x]) f[x]=_find(f[x]); return f[x]; } void _merge(int x,int y){ x=_find(x),y=_find(y); if(x!=y) f[x]=y; } void dfs(int x,int f,int dis){ vis[x]=1; if(dis>mx) mx=dis,rt=x; for(auto y:e[x]){ if(y==f) continue; dfs(y,x,dis+1); } } signed main(){ ios_base::sync_with_stdio(0); cin.tie(0),cout.tie(0); cin>>n>>m>>q; re(i,1,n) f[i]=i; while(m--){ cin>>x>>y; e[x].pub(y); e[y].pub(x); _merge(x,y); } re(i,1,n){ if(!vis[i]){ rt=0; mx=-inf; dfs(i,i,0); dfs(rt,rt,0); ans[_find(i)]=mx; } } re(i,1,q){ cin>>op; if(op==1){ cin>>x; cout<<ans[_find(x)]<<endl; } else if(op==2){ cin>>x>>y; if(_find(x)==_find(y)) continue; int tmp=mmax(ans[_find(x)],ans[_find(y)], (ans[_find(x)]+1)/2+(ans[_find(y)]+1)/2+1); _merge(x,y); ans[_find(x)]=tmp; } } return 0; } /* 10 5 10 1 2 1 3 3 4 2 5 1 6 2 7 3 1 5 2 3 4 1 3 1 8 1 8 1 8 1 9 2 5 3 2 6 5 */
实际上这题还是CF455C原题