【題目連結】

【思路要點】

• 問題等價於從起點出發只經過 $L,L+1,L+2,...,N$ 能夠到達的點和終點出發只經過 $1,2,3,...,R$ 能夠到達的點是否有交。
• 建出原圖的最小/最大生成樹的 $Kruskal $ 重構樹，這兩個點集對應了兩棵樹上的某兩個子樹。
• 對於點集根據最小生成樹的 $Kruskal $ 重構樹的 $DFS$ 序進行重標號，離線詢問，在最大生成樹的 $Kruskal $ 重構樹上線段樹合併，即可查詢子樹內是否存在標號在某一個區間中的點，從而判斷兩個點集是否有交。
• 時間複雜度 $O(NLogN)$ 。

【程式碼】

#include "werewolf.h"
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 5;
const int MAXLOG = 20;
const int MAXP = 1e7 + 5;
struct SegmentTree {
	struct Node {
		int lc, rc;
		int sum;
		bool leaf;
	} a[MAXP];
	int size, n;
	void init(
 
int x) {
		n = x;
		size = 0;
	}
	void update(int root) {
		a[root].sum = a[a[root].lc].sum + a[a[root].rc].sum;
	}
	void modify(int &root, int l, int r, int pos) {
		if (root == 0) root = ++size;
		if (l == r) {
			a[root].sum = 1;
			a[root].leaf = true;
			return;
		}
		int mid = (l + r) / 2;
		if
 
 (mid >= pos) modify(a[root].lc, l, mid, pos);
		else modify(a[root].rc, mid + 1, r, pos);
		update(root);
	}
	void modify(int &root, int val) {
		return modify(root, 1, n, val);
	}
	int merge(int x, int y) {
		if (x == 0 || y == 0) return x + y;
		if (a[x].leaf) {
			a[x].sum += a[y].sum;
			return x;
		}
		a[x].lc = merge(a[x].lc, a[y].lc);
		a[x].rc = merge(a[x].rc, a[y].rc);
		update(x);
		return x;
	}
	void join(int &to, int from) {
		to = merge(to, from);
	}
	int query(int root, int l, int r, int ql, int qr) {
		if (root == 0) return 0;
		if (l == ql && r == qr) return a[root].sum;
		int mid = (l + r) / 2, ans = 0;
		if (mid >= ql) ans += query(a[root].lc, l, mid, ql, min(mid, qr));
		if (mid + 1 <= qr) ans += query(a[root].rc, mid + 1, r, max(mid + 1, ql), qr);
		return ans;
	}
	int query(int root, int l, int r) {
		return query(root, 1, n, l, r);
	}
} ST;
vector <int> ans, a[MAXN], b[MAXN], c[MAXN];
vector <int> home[MAXN], ql[MAXN], qr[MAXN];
//a : graph, b : tree1, c : tree2.
int n, m, q, f[MAXN], root[MAXN];
int fatherb[MAXN][MAXLOG];
int fatherc[MAXN][MAXLOG];
int timer, dfn[MAXN], rit[MAXN];
int F(int x) {
	if (f[x] == x) return x;
	else return f[x] = F(f[x]);
}
void dfs(int pos, int fa) {
	fatherb[pos][0] = fa;
	for (int i = 1; i < MAXLOG; i++)
		fatherb[pos][i] = fatherb[fatherb[pos][i - 1]][i - 1];
	dfn[pos] = ++timer;
	for (unsigned i = 0; i < b[pos].size(); i++)
		dfs(b[pos][i], pos);
	rit[pos] = timer;
}
void work(int pos) {
	ST.modify(root[pos], dfn[pos]);
	for (unsigned i = 0; i < c[pos].size(); i++) {
		work(c[pos][i]);
		ST.join(root[pos], root[c[pos][i]]);
	}
	for (unsigned i = 0; i < home[pos].size(); i++)
		if (ST.query(root[pos], ql[pos][i], qr[pos][i]) != 0) ans[home[pos][i]] = 1;
}
vector <int> check_validity(int tn, vector <int> x, vector <int> y, vector <int> s, vector <int> t, vector <int> l, vector <int> r) {
	n = tn, m = x.size(), q = s.size();
	ans.resize(q);
	for (int i = 0; i <= m - 1; i++) {
		x[i]++, y[i]++;
		a[x[i]].push_back(y[i]);
		a[y[i]].push_back(x[i]);
	}
	for (int i = 1; i <= n; i++)
		f[i] = i;
	for (int i = 1; i <= n; i++) {
		for (unsigned j = 0; j < a[i].size(); j++)
			if (a[i][j] <= i && F(a[i][j]) != i) {
				b[i].push_back(F(a[i][j]));
				f[F(a[i][j])] = i;
			}
	}
	dfs(n, 0);
	for (int i = 1; i <= n; i++)
		f[i] = i;
	for (int i = n; i >= 1; i--) {
		for (unsigned j = 0; j < a[i].size(); j++)
			if (a[i][j] >= i && F(a[i][j]) != i) {
				c[i].push_back(F(a[i][j]));
				fatherc[F(a[i][j])][0] = i;
				f[F(a[i][j])] = i;
			}
	}
	for (int i = 1; i <= n; i++)
	for (int j = 1; j < MAXLOG; j++)
		fatherc[i][j] = fatherc[fatherc[i][j - 1]][j - 1];
	for (int i = 0; i <= q - 1; i++) {
		s[i]++, t[i]++;
		l[i]++, r[i]++;
		if (s[i] < l[i] || t[i] > r[i]) continue;
		int pos = t[i];
		for (int j = MAXLOG - 1; j >= 0; j--)
			if (fatherb[pos][j] != 0 && fatherb[pos][j] <= r[i]) pos = fatherb[pos][j];
		int tl = dfn[pos], tr = rit[pos];
		pos = s[i];
		for (int j = MAXLOG - 1; j >= 0; j--)
			if (fatherc[pos][j] != 0 && fatherc[pos][j] >= l[i]) pos = fatherc[pos][j];
		home[pos].push_back(i);
		ql[pos].push_back(tl);
		qr[pos].push_back(tr);
	}
	ST.init(n);
	work(1);
	return ans;
}