只看题面绝对做不出系列....

注意到\(c \leqslant 7\)，因此不会有删边操作（那样例删边干嘛）

注意到\(2, 5\)操作十分的有趣，启示我们拿线段树合并来做

操作\(7\)很好处理

操作\(6\)，维护对数的和即可

操作\(3, 4\)，乍看不好处理，然而势能分析一下就可以得出暴力的复杂度是\(O(n \log n)\)的

然而我好像写了个稳定的\(\log\)维护

然后好像就没了诶......

空间直接动态开点是开不下的....

需要预先离散化权值

复杂度\(O(n \log n)\)

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;#define de double
#define ri register int
#define rep(io, st, ed) for(ri io = st; io <= ed; 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 = 5e5 + 5;
const int eid = 5e6 + 5;de mul[eid];
int n, m, id, nc;
int rt[sid], fa[sid];
int ls[eid], rs[eid], sz[eid];
int opt[sid], c1[sid], c2[sid], T[sid];inline int find(int o) { return fa[o] = (o == fa[o]) ? o : find(fa[o]); }inline void upd(int o) {int lc = ls[o], rc = rs[o];sz[o] = sz[lc] + sz[rc];mul[o] = mul[lc] + mul[rc];
}inline int merge(int x, int y) {if(!x || !y) return x + y;ls[x] = merge(ls[x], ls[y]);rs[x] = merge(rs[x], rs[y]);sz[x] = sz[x] + sz[y];mul[x] = mul[x] + mul[y];return x;
}inline void mdf(int &o, int l, int r, int c, int v) {if(!o) o = ++ id;if(l == r) { sz[o] = v; mul[o] = (de)v * (de)log(T[c]); return; }int mid = (l + r) >> 1;if(c <= mid) mdf(ls[o], l, mid, c, v);else mdf(rs[o], mid + 1, r, c, v);upd(o);
}inline int dfs(int &o, int l, int r, int ml, int mr) {if(!o || ml > r || mr < l) return 0;if(ml <= l && mr >= r) {int tmp = sz[o]; o = 0;return tmp;}int mid = (l + r) >> 1;int ret = dfs(ls[o], l, mid, ml, mr) + dfs(rs[o], mid + 1, r, ml, mr);upd(o); return ret;
}inline int qry(int o, int l, int r, int k) {if(l == r) return T[l];int mid = (l + r) >> 1;if(sz[ls[o]] >= k) return qry(ls[o], l, mid, k);else return qry(rs[o], mid + 1, r, k - sz[ls[o]]);
}void calc() {rep(i, 1, m) {opt[i] = read(); c1[i] = read();if(opt[i] != 1 && opt[i] != 7) c2[i] = read();if(opt[i] == 1) T[++ nc] = c1[i];if(opt[i] == 3 || opt[i] == 4) T[++ nc] = c2[i];}sort(T + 1, T + nc + 1);nc = unique(T + 1, T + nc + 1) - T - 1;rep(i, 1, m) {if(opt[i] == 1) c1[i] = lower_bound(T + 1, T + nc + 1, c1[i]) - T;if(opt[i] == 3 || opt[i] == 4)c2[i] = lower_bound(T + 1, T + nc + 1, c2[i]) - T;}rep(i, 1, m) {int u, v, w, num;switch(opt[i]) {case 1 : n ++; fa[n] = n;mdf(rt[n], 1, nc, c1[i], 1); break;case 2 : u = find(c1[i]); v = find(c2[i]); if(u == v) break;fa[v] = u; rt[u] = merge(rt[u], rt[v]); break;case 3 :u = find(c1[i]); w = c2[i];num = dfs(rt[u], 1, nc, 1, w);mdf(rt[u], 1, nc, w, num); break;case 4 : u = find(c1[i]); w = c2[i];num = dfs(rt[u], 1, nc, w, nc);mdf(rt[u], 1, nc, w, num); break;case 5 :u = find(c1[i]); w = c2[i];printf("%d\n", qry(rt[u], 1, nc, w)); break;case 6 : u = find(c1[i]); v = find(c2[i]);if(mul[rt[u]] > mul[rt[v]]) puts("1");else puts("0"); break;case 7 : u = find(c1[i]);printf("%d\n", sz[rt[u]]); break;default : break;}}
}int main() {//freopen("4399.in", "r", stdin);//freopen("4399.out", "w", stdout);m = read();calc();return 0;
}