CF1140F Extending Set of Points 题解

手玩一下发现，如果我们建一个二分图，对于点集 $S$ 每个点 $(x_i,y_i)$，连左部 $x_i$ 到右部 $y_i$，那么拓展后 $|S| =$ 每个连通块的左部点数乘以右部点数。

于是我们可以愉快地线段树分治，使用可撤销并查集动态更新答案即可，复杂度线性对数平方。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
#include <map>
#include <vector>
#include <stack>
#include <utility>
using namespace std;

#define int long long

const int N = 7e5 + 5;

long long nowans = 0LL;
int n, q;

class Union_Find
{
public:
	int fa[N], sz[N], sz1[N], sz2[N];
	void Init()
	{
		for (int i = 0; i < N; i++) fa[i] = i, sz[i] = 1, sz1[i] = (i <= 300000), sz2[i] = (i > 300000);
	}
	int find(int u)
	{
		return (fa[u] == u ? u : find(fa[u]));
	}
	pair<int, int> merge(int u, int v)
	{
		if ((u = find(u)) == (v = find(v))) return make_pair(-1, -1);
		if (sz[u] < sz[v]) swap(u, v);
		fa[v] = u;
		nowans -= 1LL * sz1[u] * sz2[u] + 1LL * sz1[v] * sz2[v];
		sz[u] += sz[v];
		sz1[u] += sz1[v], sz2[u] += sz2[v];
		nowans += 1LL * sz1[u] * sz2[u];
		return make_pair(u, v);
	}
	void del(int u, int v)
	{
		sz[u] -= sz[v];
		nowans -= 1LL * sz1[u] * sz2[u];
		sz1[u] -= sz1[v], sz2[u] -= sz2[v];
		nowans += 1LL * sz1[u] * sz2[u] + 1LL * sz1[v] * sz2[v];
		fa[v] = v;
	}
}uf;

class SegmentTree
{
public:
	struct Node
	{
		int l, r;
		vector<pair<int, int>> v;
	}tr[N << 2];
	void build(int u, int l, int r)
	{
		tr[u] = { l, r };
		tr[u].v.clear(), tr[u].v.shrink_to_fit();
		if (l == r) return;
		int mid = l + r >> 1;
		build(u << 1, l, mid);
		build(u << 1 | 1, mid + 1, r);
	}
	void update(int u, int l, int r, auto v)
	{
		if (tr[u].l >= l and tr[u].r <= r)
		{
			tr[u].v.emplace_back(v);
			return;
		}
		int mid = tr[u].l + tr[u].r >> 1;
		if (l <= mid) update(u << 1, l, r, v);
		if (r > mid) update(u << 1 | 1, l, r, v);
	}
	void solve(int u)
	{
		stack<pair<int, int>> st;
		for (auto& j : tr[u].v)
		{
			auto p = uf.merge(j.first, j.second + 300000);
			if (p.first != -1) st.push(p);
		}
		if (tr[u].l == tr[u].r)
		{
			cout << nowans << " ";
		}
		else
		{
			solve(u << 1);
			solve(u << 1 | 1);
		}
		while (st.size())
		{
			uf.del(st.top().first, st.top().second);
			st.pop();
		}
	}
}sgt;

map<pair<int, int>, int> mp;

signed main()
{
	ios::sync_with_stdio(0), cin.tie(0);
	cin >> q;
	sgt.build(1, 1, q);
	for (int i = 1; i <= q; i++)
	{
		int x, y;
		cin >> x >> y;
		if (mp[make_pair(x, y)])
		{
			sgt.update(1, mp[make_pair(x, y)], i - 1, make_pair(x, y));
			//cout << "!!!!!: " << mp[make_pair(x, y)] << " " << i - 1 << " " << x << " " << y << "\n";
			mp[make_pair(x, y)] = 0;
		}
		else mp[make_pair(x, y)] = i;
	}
	for (auto& [x, y] : mp)
	{
		if (y <= q && y >= 1)
		{
			sgt.update(1, y, q, make_pair(x.first, x.second));
			//cout << "!!!!!: " << y << " " << q << " " << x.first << " " << x.second << "\n";
		}
	}
	uf.Init();
	sgt.solve(1);
	return 0;
}