AT2336 [ARC069D] Flags

AT2336 [ARC069D] Flags

二分答案，2-sat 判定，线段树优化建边。

线段树上父亲向儿子连边。

对于每个点 $x$，连向与他距离不超过 $mid$ 的点的反点，表示选 $x$ 就只能选与他距离不超过 $mid$ 的点。

最后跑 tarjan 判定即可。

时间复杂度 $\mathcal{O}(n{\log }^{2}n)$。

#include<bits/stdc++.h>

using namespace std;

//#define int long long
typedef long long ll;

#define ha putchar(' ')
#define he putchar('\n')

inline int read() {
    
	int x = 0, f = 1;
	char c = getchar();
	while (c < '0' || c > '9') {
    
		if (c == '-')
			f = -1;
		c = getchar();
	}
	while (c >= '0' && c <= '9')
		x = x * 10 + c - '0', c = getchar();
	return x * f;
}

inline void write(int x) {
    
	if (x < 0) {
    
		putchar('-');
		x = -x;
	}
	if (x > 9)
		write(x / 10);
	putchar(x % 10 + '0');
}

const int _ = 8e4 + 10;

int n, cnt, idx, scc, low[_], dfn[_], id[_], Id[_];

stack<int> s;

vector<pair<int, int>> d;

vector<int> e[_];

void build(int o, int l, int r) {
    
	id[o] = ++cnt;
	if (o > 1) e[id[o >> 1]].emplace_back(cnt);
	if (l == r) {
    
		int v = d[l - 1].second;
		e[id[o]].push_back(v <= n ? v + n : v - n);
		return;
	}
	int mid = (l + r) >> 1;
	build(o << 1, l, mid), build(o << 1 | 1, mid + 1, r);
}

void upd(int o, int l, int r, int L, int R, int x) {
    
	if(L > R) return;
	if (L <= l && r <= R) {
    
		e[x].emplace_back(id[o]);
		return;
	}
	int mid = (l + r) >> 1;
	if (L <= mid) upd(o << 1, l, mid, L, R, x);
	if (R > mid) upd(o << 1 | 1, mid + 1, r, L, R, x);
}

pair<int, int> get(int i, int m) {
    
	pair<int, int> ret;
	int l = 1, r = i, mid;
	while (l <= r) {
    
		mid = (l + r) >> 1;
		if (d[i - 1].first - d[mid - 1].first >= m) l = mid + 1;
		else r = mid - 1;
	}
	ret.first = r + 1, l = i, r = n << 1;
	while (l <= r) {
    
		mid = (l + r) >> 1;
		if (d[mid - 1].first - d[i - 1].first < m) l = mid + 1;
		else r = mid - 1;
	}
	ret.second = l - 1;
	return ret;
}

void tarjan(int u) {
    
	low[u] = dfn[u] = ++idx;
	s.push(u);
	for (int v : e[u])
		if (!dfn[v]) {
    
			tarjan(v);
			low[u] = min(low[u], low[v]);
		} else if (!Id[v]) low[u] = min(low[u], dfn[v]);
	if (low[u] == dfn[u]) {
    
		++scc;
		while (1) {
    
			int nw = s.top();
			s.pop();
			Id[nw] = scc;
			if (nw == u) break;
		}
	}
}

bool check(int lim) {
    
	scc = idx = 0;
	memset(low, 0, sizeof low);
	memset(dfn, 0, sizeof dfn);
	memset(Id, 0, sizeof Id);
	for (int i = 0; i < _; ++i) e[i].clear();
	build(1, 1, cnt = n << 1);
	for (int i = 1; i <= n << 1; ++i) {
    
		int r = d[i - 1].second;
		pair<int, int> p = get(i, lim);
		upd(1, 1, n << 1, p.first, i - 1, r);
		upd(1, 1, n << 1, i + 1, p.second, r);
	}
	for (int i = 1; i <= n << 1; ++i)
		if (!dfn[i]) tarjan(i);
	for (int i = 1; i <= n; ++i)
		if (Id[i] == Id[i + n]) return 0;
	return 1;
}

signed main() {
    
	n = read();
	for (int i = 1, x, y; i <= n; ++i) {
    
		x = read(), y = read();
		d.emplace_back(make_pair(x, i));
		d.emplace_back(make_pair(y, i + n));
	}
	sort(d.begin(), d.end());
	int l = 0, r = 1e9, mid;
	while (l <= r) {
    
		mid = (l + r) >> 1;
		if (check(mid)) l = mid + 1;
		else r = mid - 1;
	}
	write(l - 1), he;
	return 0;
}