Codeforces 1140F 线段树 分治 并查集

题意及思路：https://blog.csdn.net/u013534123/article/details/89010251

之前cf有一个和这个相似的题，不过那个题只有合并操作，没有删除操作，直接并查集搞一搞就行了。对于这个题，因为有删除操作，我们对操作序列建一颗线段树，记录每个操作影响的区间操作就可以了。这里的并查集不能路径压缩，要按秩合并，这样复杂度是O(logn)的。

代码：

#include <bits/stdc++.h>
#define ls (o << 1)
#define rs (o << 1 | 1)
#define INF 0x3f3f3f3f
#define db double
#define pii pair<int, int>
#define LL long long
using namespace std;
const int maxn = 300010;
const int Base = 300000;
vector<pii> tr[maxn * 4];
map<pii, int> mp;
map<pii, int>::iterator it;
LL res[maxn], cnt_x[maxn * 2], cnt_y[maxn * 2], sz[maxn * 2];
int f[maxn * 2];
LL ans;
pii a[maxn];
int get(int x) {
	if(x == f[x]) return x;
	return get(f[x]);
}
void add(int o, int l, int r, int ql, int qr, pii val) {
	if(l >= ql &&r <= qr) {
		tr[o].push_back(val);
		return;
	}
	int mid = (l + r) >> 1;
	if(ql <= mid) add(ls, l, mid, ql, qr, val);
	if(qr > mid) add(rs, mid + 1, r, ql, qr, val);
}
void del(int x, int y) {
	int x1 = get(x), y1 = get(y);
	if(x1 != y1) return;
	ans -= cnt_x[x] * cnt_y[x];
	cnt_x[x] -= cnt_x[y], cnt_y[x] -= cnt_y[y];
	sz[x] -= sz[y];
	ans += cnt_x[x] * cnt_y[x];
	ans += cnt_x[y] * cnt_y[y];
	f[y] = y;
}
pii merge(int x, int y) {
	int x1 = get(x), y1 = get(y);
	if(x1 == y1) return make_pair(-1, -1);
	if(sz[x1] < sz[y1]) swap(x1, y1);
	ans -= cnt_x[x1] * cnt_y[x1];
	ans -= cnt_x[y1] * cnt_y[y1];
	sz[x1] += sz[y1];
	cnt_x[x1] += cnt_x[y1], cnt_y[x1] += cnt_y[y1];
	ans += cnt_x[x1] * cnt_y[x1];
	f[y1] = x1;
	return make_pair(x1, y1);
}
void dfs(int o, int l, int r) {
	if(l == 12) {
		l++;
		l--;
	}
	stack<pii> s;
	for (auto x : tr[o]) {
		pii tmp = merge(x.first, x.second);
		if(tmp.first != -1) s.push(tmp);
	}
	if(l == r) res[l] = ans;
	else {
		int mid = (l + r) >> 1;
		dfs(ls, l, mid);
		dfs(rs, mid + 1, r);
	}
	while(!s.empty()) {
		del(s.top().first, s.top().second);
		s.pop();
	}
}
int main() {
	int n, x, y;
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%d%d", &x, &y);
		y += Base;
		a[i] = make_pair(x, y);
		if(mp.find(a[i]) == mp.end()) mp[a[i]] = i;
		else {
			add(1, 1, n, mp[a[i]], i - 1, a[i]);
			mp.erase(a[i]);
		}
	}
	for (it = mp.begin(); it != mp.end(); it++) {
		add(1, 1, n, it -> second, n, it -> first);
	}
	for (int i = 1; i <= Base; i++) {
		f[i] = i, cnt_x[i] = 1, cnt_y[i] = 0, sz[i] = 1;
	}
	for (int i = Base + 1; i <= Base * 2; i++) {
		f[i] = i, cnt_x[i] = 0, cnt_y[i] = 1, sz[i] = 1;
	}
	dfs(1, 1, n);
	for (int i = 1; i <= n; i++)
		printf("%lld ", res[i]);
}

原文地址：https://www.cnblogs.com/pkgunboat/p/11027129.html