考虑\(LCT\)

不难发现，我们不需要换根...

对于操作\(1\)，\(splay(u)\)然后连虚边即可

对于操作\(3\)，我们可以先\(access(u)\)，然后再\(access(v)\)，然后查最后一个虚边变实边的点

对于操作\(2\)

可以选择\(access(u), splay(u)\)，然后从\(u\)所在的\(splay\)中删去\(u\)点

也可以选择\(access(u), access(v), splay(u)\)，这时，边\((u, v)\)成为虚边，十分好删除

复杂度\(O(n \log n)\)

版本1：

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;#define ri register int
#define rep(io, st, ed) for(ri io = st; io <= ed; io ++)
#define drep(io, ed, st) for(ri io = ed; io >= st; io --)#define gc getchar
inline int read() {int p = 0, w = 1; char c = gc();while(c > '9' || c < '0') { if(c == '-') w = -1; c = gc(); }while(c >= '0' && c <= '9') p = p * 10 + c - '0', c = gc();return p * w;
}const int sid = 1e5 + 5;int n, m;
char s[sid];
int son[sid][2], fa[sid], pra[sid];#define ls(o) son[(o)][0]
#define rs(o) son[(o)][1]inline bool isrc(int o) { return rs(fa[o]) == o; }
inline bool isr(int o) { return !fa[o] || (ls(fa[o]) != o && rs(fa[o]) != o); }inline void rotate(int o) {int f = fa[o], g = fa[f];int ro = isrc(o), rf = isrc(f), p = son[o][ro ^ 1];if(!isr(f)) son[g][rf] = o; son[o][ro ^ 1] = f; son[f][ro] = p;fa[p] = f; fa[f] = o; fa[o] = g;
}inline void splay(int o) {while(!isr(o)) {int f = fa[o];if(!isr(f)) rotate(isrc(f) == isrc(o) ? f : o);rotate(o);}
}int lca = 0;
inline void access(int o) {int lst = 0;while(o) {splay(o); rs(o) = lst;lca = lst = o; o = fa[o];}
}int main() {n = read(); m = read();rep(i, 1, m) {int u, v;scanf("%s", s);if(s[1] == 'i') {u = read(); v = read();splay(u); pra[u] = v; fa[u] = v;}else if(s[1] == 'c') {u = read(); v = read();access(u); access(v);printf("%d\n", lca);}else if(s[1] == 'u') {u = read();access(u); access(pra[u]);splay(u); fa[u] = 0;}}return 0;
}

版本\(2\)：

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;#define ri register int
#define rep(io, st, ed) for(ri io = st; io <= ed; io ++)
#define drep(io, ed, st) for(ri io = ed; io >= st; io --)#define gc getchar
inline int read() {int p = 0, w = 1; char c = gc();while(c > '9' || c < '0') { if(c == '-') w = -1; c = gc(); }while(c >= '0' && c <= '9') p = p * 10 + c - '0', c = gc();return p * w;
}const int sid = 1e5 + 5;int n, m;
char s[sid];
int son[sid][2], fa[sid], pra[sid];#define ls(o) son[(o)][0]
#define rs(o) son[(o)][1]inline bool isrc(int o) { return rs(fa[o]) == o; }
inline bool isr(int o) { return !fa[o] || (ls(fa[o]) != o && rs(fa[o]) != o); }inline void rotate(int o) {int f = fa[o], g = fa[f];int ro = isrc(o), rf = isrc(f), p = son[o][ro ^ 1];if(!isr(f)) son[g][rf] = o; son[o][ro ^ 1] = f; son[f][ro] = p;fa[p] = f; fa[f] = o; fa[o] = g;
}inline void splay(int o) {while(!isr(o)) {int f = fa[o];if(!isr(f)) rotate(isrc(f) == isrc(o) ? f : o);rotate(o);}
}int lca = 0;
inline void access(int o) {int lst = 0;while(o) {splay(o); rs(o) = lst;lca = lst = o; o = fa[o];}
}int main() {n = read(); m = read();rep(i, 1, m) {int u, v;scanf("%s", s);if(s[1] == 'i') {u = read(); v = read();splay(u); pra[u] = v; fa[u] = v;}else if(s[1] == 'c') {u = read(); v = read();access(u); access(v);printf("%d\n", lca);}else if(s[1] == 'u') {u = read();access(u); splay(u);ls(u) = fa[ls(u)] = 0;}}return 0;
}