P7561 [JOISC 2021 Day2] 道路の建設案

题意

给定 \(n\) 个点，求平面上，曼哈顿距离最近的 \(k\) 点对。

Sol

仔细想想就会发现，曼哈顿距离不好做最近 \(k\) 点对。

考虑转成切比雪夫距离。\(x' = x + y, y' = x - y\)。

二分答案，每次 \(check\) 一个 \(dis\)，询问距离小于 \(dis\) 的点对是否有 \(k\) 个。

\(check\) 是平凡的，用尺取做掉一维数据结构维护另一维即可。

Code

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <array>
#include <set>
#include <cassert>
#define int long long
#define pii pair <int, int>
using namespace std;
#ifdef ONLINE_JUDGE

#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
char buf[1 << 23], *p1 = buf, *p2 = buf, ubuf[1 << 23], *u = ubuf;

#endif
int read() {
	int p = 0, flg = 1;
	char c = getchar();
	while (c < '0' || c > '9') {
		if (c == '-') flg = -1;
		c = getchar();
	}
	while (c >= '0' && c <= '9') {
		p = p * 10 + c - '0';
		c = getchar();
	}
	return p * flg;
}
void write(int x) {
	if (x < 0) {
		x = -x;
		putchar('-');
	}
	if (x > 9) {
		write(x / 10);
	}
	putchar(x % 10 + '0');
}
#define fi first
#define se second

const int N = 2.5e5 + 5, inf = 1e10;

set <pii> isl;
array <pii, N> s;
array <int, N> cur;

int cnt;

bool check(int x, int n, int k) {
	int rd = 1;
	isl.clear(); cnt = 0;
	for (int i = 1; i <= n; i++) {
		rd = max(rd, i);
		isl.erase(make_pair(s[i].se, s[i].fi));
		while (rd < n && s[rd + 1].fi - s[i].fi <= x) {
			rd++;
			isl.insert(make_pair(s[rd].se, s[rd].fi));
		}
		auto it = isl.lower_bound(make_pair(s[i].se - x, -inf));
		/* if (i == 1 && it != isl.end()) { */
			/* write(s[i].fi), putchar(32); */
			/* write(s[i].se), puts(""); */
		/* } */
		for (; it != isl.end(); it++) {
			int tp = max(abs(it -> se - s[i].fi), abs(it -> fi - s[i].se));
			if (tp > x) break;
			cnt++, cur[cnt] = tp;
			if (cnt >= k) return true;
		}
	}
	return false;
}

signed main() {
	int n = read(), k = read();
	for (int i = 1; i <= n; i++) {
		int x = read(), y = read();
		s[i] = make_pair(x + y, x - y);
	}

	sort(s.begin() + 1, s.begin() + 1 + n);
	int l = 1, r = 1e10 + 5;
	int ans = -1;
	while (l < r) {
		int mid = (l + r) >> 1;
		if (check(mid, n, k)) ans = r = mid;
		else l = mid + 1;
	}
	check(ans - 1, n, k);
	sort(cur.begin() + 1, cur.begin() + 1 + cnt);
	for (int i = 1; i <= cnt; i++)
		write(cur[i]), puts("");
	for (int i = cnt + 1; i <= k; i++)
		write(ans), puts("");
	return 0;
}