P3224 [HNOI2012] 永无乡

题意

思路

用并查集维护连通性，每个集合维护一个平衡树，每次合并两个集合的时候，将一个平衡树的节点一个一个加入到另一个中。

这么做不会超时，每次将小的平衡树拆掉放到大的中，可以证明不会超过 \(O(\log n)\) 次。

总时间复杂度 \(O(n \log ^ 2 n)\)。

代码

#include <bits/stdc++.h>using namespace std;const int N = 1000010;struct node {int l, r;int size;int rnd;int key;int x;
} tr[N];int rt[N], idx;void pushup(int u) {tr[u].size = tr[tr[u].l].size + tr[tr[u].r].size + 1;
}void split(int u, int key, int& x, int& y) {if (!u) {x = y = 0;return;}if (tr[u].key <= key) {x = u;split(tr[u].r, key, tr[u].r, y);}else {y = u;split(tr[u].l, key, x, tr[u].l);}pushup(u);
}int merge(int x, int y) {if ((!x) || (!y)) return x | y;if (tr[x].rnd < tr[y].rnd) {tr[x].r = merge(tr[x].r, y);pushup(x);return x;}else {tr[y].l = merge(x, tr[y].l);pushup(y);return y;}
}int newnode(int key, int x) {idx++;tr[idx].key = key;tr[idx].rnd = rand();tr[idx].size = 1;tr[idx].l = tr[idx].r = 0;tr[idx].x = x;return idx;
}void insert(int& rt, int key, int g) {int x, y;split(rt, key, x, y);int z = newnode(key, g);rt = merge(merge(x, z), y);
}int fa[N];int find(int x) {if (x == fa[x]) return x;return fa[x] = find(fa[x]);
}void merge_fa(int x, int y) {x = find(x), y = find(y);if (x == y) return;if (tr[rt[x]].size > tr[rt[y]].size) swap(x, y);fa[x] = y;queue<int> q;q.push(rt[x]);while (q.size()) {int t = q.front();q.pop();insert(rt[y], tr[t].key, tr[t].x);if (tr[t].l) q.push(tr[t].l);if (tr[t].r) q.push(tr[t].r);}
}int get_rank(int u, int rk) {if (tr[tr[u].l].size + 1 == rk) return tr[u].x;if (tr[tr[u].l].size >= rk) return get_rank(tr[u].l, rk);return get_rank(tr[u].r, rk - tr[tr[u].l].size - 1);
}int n, m;int main() {ios::sync_with_stdio(false);cin.tie(nullptr);cin >> n >> m;for (int i = 1; i <= n; i++) {int x;cin >> x;insert(rt[i], x, i);fa[i] = i;}for (int i = 1; i <= m; i++) {int u, v;cin >> u >> v;merge_fa(u, v);}int q;cin >> q;char opt;int x, y;while (q--) {cin >> opt >> x >> y;if (opt == 'Q') {x = find(x);if (tr[rt[x]].size < y) cout << "-1\n";else cout << get_rank(rt[x], y) << '\n';}else merge_fa(x, y);}return 0;
}