[ARC069F] Flags

题目

思路

我们一定能找到唯一的一个分界点 \(x\) 使得间隔 \(x - 1\) 不存在解，而 \(x\) 存在解。

考虑二分，然后 2-SAT 建图进行 check。

建图是 2-SAT 的经典套路，就不说了。

代码

#include <bits/stdc++.h>

using namespace std;
using PII = pair<int, int>;

const int N = 10000010, M = 80000010;

int n;
int dfn[N], low[N], cnt;
int stk[N], top;
bool instk[N];
int scc[N], tot;
PII p[N];

struct edge {
    int to, next;
} e[M];

int head[N], idx;

void add(int u, int v) {
    idx++, e[idx].to = v, e[idx].next = head[u], head[u] = idx;
}

struct node {
    int l, r;
} tr[N * 2];

int segtot, root;

void build(int& u, int l, int r) {
    u = ++segtot;
    if (l == r) {
        add(u, p[l].second <= n ? p[l].second + n : p[l].second - n);
        return;
    }
    int mid = l + r >> 1;
    build(tr[u].l, l, mid);
    add(u, tr[u].l);
    build(tr[u].r, mid + 1, r);
    add(u, tr[u].r);
}

void connect(int u, int l, int r, int pl, int pr, int v) {
    if (pl <= l && r <= pr) {
        add(v, u);
        return;
    }
    int mid = l + r >> 1;
    if (pl <= mid) connect(tr[u].l, l, mid, pl, pr, v);
    if (pr > mid) connect(tr[u].r, mid + 1, r, pl, pr, v);
}

void tarjan(int u) {
    stk[++top] = u, instk[u] = 1;
    dfn[u] = low[u] = ++cnt;
    for (int i = head[u]; i; i = e[i].next) {
        int to = e[i].to;
        if (!dfn[to]) {
            tarjan(to);
            low[u] = min(low[u], low[to]);
        }
        else if (instk[to]) low[u] = min(low[u], dfn[to]);
    }

    if (dfn[u] == low[u]) {
        tot++;
        while (stk[top] != u) {
            int t = stk[top];
            top--;
            instk[t] = 0;
            scc[t] = tot;
        }
        top--;
        instk[u] = 0;
        scc[u] = tot;
    }
}


bool check(int t) {
    memset(head, 0, sizeof(head));
    memset(dfn, 0, sizeof(dfn));
    memset(instk, 0, sizeof(instk));
    memset(scc, 0, sizeof(scc));
    root = cnt = tot = top = idx = 0;
    segtot = 2 * n;

    build(root, 1, 2 * n);

    for (int i = 1; i <= 2 * n; i++) {
        int x = upper_bound(p + 1, p + 2 * n + 1, make_pair(p[i].first - t, 1000000000)) - p;
        int y = lower_bound(p + 1, p + 2 * n + 1, make_pair(p[i].first + t, -1000000000)) - p - 1;
        if (x < i) connect(root, 1, 2 * n, x, i - 1, p[i].second);
        if (i < y) connect(root, 1, 2 * n, i + 1, y, p[i].second);
    }
    
    for (int i = 1; i <= 2 * n; i++) if (!dfn[i]) tarjan(i);
    
    for (int i = 1; i <= n; i++) {
        if (scc[i] == scc[i + n]) {
            return 0;
        }
    }
    return 1;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> p[i].first >> p[i + n].first;
        p[i].second = i, p[i + n].second = i + n;
    }

    sort(p + 1, p + 2 * n + 1);

    int l = 0, r = p[2 * n].first - p[1].first + 1;
    while (l < r) {
        int mid = l + r + 1 >> 1;
        if (check(mid)) l = mid;
        else r = mid - 1;
    }
    cout << l << '\n';
    return 0;
}