线段树分治

维护问题有这个特征：数据在 \([l, r]\) 时刻出现。

常见的是对时间建线段树。具体做法是遍历整颗线段树，到达每个节点时执行相应的操作，然后继续向下递归，到达叶子节点时统计贡献，回溯时撤销操作即可。

Luogu P5787 二分图 /【模板】线段树分治

考虑什么情况是二分图：充要条件是不存在奇环，那么就可以维护一个可撤销并查集（用栈）维护即可。

#include <bits/stdc++.h>
#define int long long 

using namespace std;

const int N = 4e5 + 100;

int n, m, k, dep[N], fa[N], top, ans[N];

vector<pair<int, int> > g[N];

struct Node {
	int x, y, high;
} st[N];

void insert(int p, int l, int r, int ql, int qr, pair<int, int> x) {
	if (ql <= l && qr >= r) {
		g[p].push_back(x);
		return ;
	}
	int mid = (l + r) >> 1;
	if (ql <= mid) {
		insert(p << 1, l, mid, ql, qr, x);
	}
	if (qr > mid) {
		insert(p << 1 | 1, mid + 1, r, ql, qr, x);
	}
}

int getfa(int x) {
	return x == fa[x] ? x : getfa(fa[x]);
}

void merge(int x, int y) {
	x = getfa(x);
	y = getfa(y);
	if (dep[x] > dep[y]) {
		swap(x, y);
	}
	st[++top] = {x, y, dep[y]};
	fa[x] = y;
	dep[y] += (dep[x] == dep[y]);
}

void solve(int p, int l, int r) {
	bool flag = 0;
	int now = top;
	for (auto v : g[p]) {
		merge(v.first, v.second + n);
		merge(v.second, v.first + n);
		if (getfa(v.first) == getfa(v.second)) {
			flag = 1;
			break;
		}
	}
	if (!flag) {
		if (l == r) {
			ans[l] = 1;
		} else {
			int mid = (l + r) >> 1;
			solve(p << 1, l, mid);
			solve(p << 1 | 1, mid + 1, r);
		}
	}
	while (top > now) {
		Node t = st[top--];
		fa[t.x] = t.x;
		dep[t.y] = t.high;
	}
}

signed main() {
	cin >> n >> m >> k;
	for (int i = 1; i <= 2 * n; i++) {
		fa[i] = i;
	}
	for (int i = 1, x, y, l, r; i <= m; i++) {
		cin >> x >> y >> l >> r;
		insert(1, 1, k, l + 1, r, {x, y});
	}
	solve(1, 1, k);
	for (int i = 1; i <= k; i++) {
		cout << (ans[i] ? "Yes\n" : "No\n");
	}
	return 0;
}

Luogu P5631 最小 mex 生成树

考虑数 \(x\) 在什么样的情况下会成为 \(mex\)：\(1\sim \text{x} - 1\) 出现，但是 \(x\) 不出现，那我们就可以对每一个 \(w_i\) 维护线段树：

insert(1, 0, mx, 0, w[i] - 1, i);
insert(1, 0, mx, w[i] + 1, mx, i);

#include <bits/stdc++.h>
#define int long long 

using namespace std;

const int N = 4e6 + 100;

int n, m, siz[N], fa[N], top, a[N], b[N], w[N];

vector<int> g[N];

struct Node {
	int x, y, siz;
} st[N];

void insert(int p, int l, int r, int ql, int qr, int x) {
	if (ql > r || qr < l) {
		return ;
	}
	if (ql <= l && qr >= r) {
		g[p].push_back(x);
		return ;
	}
	int mid = (l + r) >> 1;
	insert(p << 1, l, mid, ql, qr, x);
	insert(p << 1 | 1, mid + 1, r, ql, qr, x);
}

int getfa(int x) {
	return x == fa[x] ? x : getfa(fa[x]);
}

void merge(int x, int y) {
	x = getfa(x);
	y = getfa(y);
	if (x == y) {
		return ;
	}
	if (siz[x] > siz[y]) {
		swap(x, y);
	}
	st[++top] = {x, y, siz[y]};
	fa[x] = y;
	siz[y] += siz[x];
}

void solve(int p, int l, int r) {
	int now = top;
	for (int i : g[p]) {
		merge(a[i], b[i]);
	}
//	cout << "(p, l, r) = (" << p << ", " << l << ", " << r << ")\n";
	if (l == r) {
//		cout << "siz = " << siz[getfa(1)] << '\n';
		if (siz[getfa(1)] == n) {
			cout << l << '\n';
			exit(0);
		}
	} else {
		int mid = (l + r) >> 1;
		solve(p << 1, l, mid);
		solve(p << 1 | 1, mid + 1, r);
	}
	while (top > now) {
		Node t = st[top--];
		fa[t.x] = t.x;
		siz[t.y] = t.siz;
	}
}

signed main() {
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
//	freopen("1.in", "r", stdin);
	cin >> n >> m;
	int mx = 0, mn = 1e9;
	for (int i = 1; i < N; i++) {
		fa[i] = i, siz[i] = 1;
	}
	for (int i = 1; i <= m; i++) {
		cin >> a[i] >> b[i] >> w[i];
		mx = max(mx, w[i]);
		mn = min(mn, w[i]);
	}
	mx++;
	if (mn > 0) {
		cout << "0\n";
		return 0;
	}
	for (int i = 1; i <= m; i++) {
		insert(1, 0, mx, 0, w[i] - 1, i);
		insert(1, 0, mx, w[i] + 1, mx, i);
	}
	solve(1, 0, mx);
	return 0;
}