平面最近點對 & 最小周長三角形 & 曼哈頓距離最近

Laijinyi發表於2024-10-07

Statement

求平面最近點對的距離，距離定義為歐幾里德距離。

Solution

考慮按 \(x\) 排序，分治計算

先往左右遞迴，設得到的答案為 \(d\)，當前算跨過中點的答案，那麼答案 \(\ge d\) 的點對可以不用列舉

設中點為 \(m\)

對 \(i\in[l..m]\)，\(x_i\le x_m-d\) 的不用列舉；對 \(j\in[m+1..r]\)，\(x_j\ge x_m+d\) 的不用列舉
對於所有 \(i\in[l..m]\)，不超過 6 個 \(j\in[m+1..r]\) 滿足 \(x_j\in[x_m,x_m+d]\) 且 \(y_j\in[y_i-d,y_i+d]\)

於是對於需要列舉的點，按 \(y\) 排序，左邊每個 \(i\) 列舉不超過 \(6\) 個候選點更新答案，\(O(n\log n)\)

Code

#include <bits/stdc++.h>
using namespace std;
#define rep(i, j, k) for (int i = (j); i <= (k); ++i)
#define reo(i, j, k) for (int i = (j); i >= (k); --i)
typedef long long ll;
const int N = 5e5 + 10;
struct Point {
	ll x, y;
} a[N], b[N];
int n;
ll ans;

ll dis(int u, int v) {
	return (a[u].x - a[v].x) * (a[u].x - a[v].x) + (a[u].y - a[v].y) * (a[u].y - a[v].y);
}
ll dis(Point u, Point v) {
	return (u.x - v.x) * (u.x - v.x) + (u.y - v.y) * (u.y - v.y);
}
void Solve(int l, int r) {
	if (r - l < 3) {
		rep(i, l, r) rep(j, i + 1, r) ans = min(ans, dis(i, j));
		sort(a + l, a + r + 1, [&](const Point& u, const Point& v) {
			return u.y < v.y;
		});
		return;
	}
	int mid = (l + r) >> 1;
	ll X = a[mid].x;
	Solve(l, mid), Solve(mid + 1, r);
	ll d = ceil(sqrt(ans));
	vector<Point> L, R;
	rep(i, l, mid)
		if (a[i].x > X - d) L.push_back(a[i]);
	rep(i, mid + 1, r)
		if (a[i].x < X + d) R.push_back(a[i]);
	int sz = R.size(), pl = 0, pr = 0;
	if (sz) {
		for (auto p : L) {
			while (pl < sz && R[pl].y < p.y - d) ++pl;
			while (pr < sz && R[pr].y <= p.y + d) ++pr;
			pr = max(pr - 1, 0);
			rep(i, pl, pr) ans = min(ans, dis(p, R[i]));
		}
	}
	for (int i = l, j = mid + 1, tot = l; i <= mid || j <= r; )
		if (i <= mid && (j > r || a[i].y <= a[j].y)) b[tot++] = a[i++];
		else b[tot++] = a[j++];
	rep(i, l, r) a[i] = b[i];
}

int main() {
	ios::sync_with_stdio(false), cin.tie(nullptr);
	cin >> n;
	rep(i, 1, n) {
		cin >> a[i].x >> a[i].y;
	}
	sort(a + 1, a + n + 1, [&](const Point& u, const Point& v) {
		return u.x < v.x;
	});
	ans = 1e18;
	Solve(1, n);
	cout << ans << '\n';
	return 0;
}

Statement

找出最小周長三角形的周長。為減小難度，這裡三角形也包含共線的三點。

Solution

這裡是距離 \(\ge\frac d2\) 的不用列舉，注意到這裡候選點數同樣是常數。

Code

#include <bits/stdc++.h>
using namespace std;
#define rep(i, j, k) for (int i = (j); i <= (k); ++i)
#define reo(i, j, k) for (int i = (j); i >= (k); --i)
typedef long long ll;
const int N = 2e5 + 10;
struct Point {
	double x, y;
} a[N];
int n;
double ans = 1e18;

double dis(int u, int v) {
	return sqrt(1.0 * (a[u].x - a[v].x) * (a[u].x - a[v].x) + 1.0 * (a[u].y - a[v].y) * (a[u].y - a[v].y));
}
double dis(Point u, Point v) {
	return sqrt(1.0 * (u.x - v.x) * (u.x - v.x) + 1.0 * (u.y - v.y) * (u.y - v.y));
}
void Solve(int l, int r) {
	if (r - l < 3) {
		if (r - l == 2) 
			ans = min(ans, dis(l, l + 1) + dis(l, l + 2) + dis(l + 1, l + 2));
		return;
	}
	int mid = (l + r) >> 1;
	Solve(l, mid), Solve(mid + 1, r);
	double d = ans / 2.0;
	vector<Point> L, R;
	rep(i, l, mid) 
		if (a[i].x >= a[mid].x - d) L.push_back(a[i]);
	rep(i, mid + 1, r)
		if (a[i].x <= a[mid].x + d) R.push_back(a[i]);
	sort(L.begin(), L.end(), [&](const Point& u, const Point& v) {
		return u.y < v.y;
	});
	sort(R.begin(), R.end(), [&](const Point& u, const Point& v) {
		return u.y < v.y;
	});
	int sz = R.size(), pl = 0, pr = 0;
	for (auto i : L) {
		while (pl < sz && R[pl].y < i.y - d) ++pl;
		while (pr < sz && R[pr].y <= i.y + d) ++pr;
		pr = max(0, pr - 1);
		if (pr - pl + 1 >= 2) 
			rep(u, pl, pr)
				rep(v, u + 1, pr) {
					ans = min(ans, dis(i, R[u]) + dis(i, R[v]) + dis(R[u], R[v]));
				}
	}
	sz = L.size(), pl = 0, pr = 0;
	for (auto i : R) {
		while (pl < sz && L[pl].y < i.y - d) ++pl;
		while (pr < sz && L[pr].y <= i.y + d) ++pr;
		pr = max(0, pr - 1);
		if (pr - pl + 1 >= 2) 
			rep(u, pl, pr)
				rep(v, u + 1, pr) {
					ans = min(ans, dis(i, L[u]) + dis(i, L[v]) + dis(L[u], L[v]));
				}
	}
}

int main() {
	ios::sync_with_stdio(false), cin.tie(nullptr);
	cin >> n;
	rep(i, 1, n) {
		cin >> a[i].x >> a[i].y;
	}
	sort(a + 1, a + n + 1, [&](const Point& u, const Point& v) {
		return u.x < v.x;
	});
	Solve(1, n);
	cout << fixed << setprecision(6) << ans << '\n';
	return 0;
}

Statement

加點
求曼哈頓距離與 \(u\) 最近的距離

Solution

問題等價於三維偏序

因為若 \(x_i\ge x,y_i\ge y\)，\(\min_{i}\{|x_i-x|+|y_i-y|\}=\min\{x_i+y_i-(x+y)\}=\min\{x_i+y_i\}-(x+y)\)

Code

#include <bits/stdc++.h>
using namespace std;
#define rep(i, j, k) for (int i = (j); i <= (k); ++i)
#define reo(i, j, k) for (int i = (j); i >= (k); --i)
typedef long long ll;
const int N = 3e6 + 10;
struct Oper {
	int op, x, y, id;
} a[N], b[N], c[N];
int n, m, tot, qtot, ans[N];

struct BIT {
	int mn[N];
	BIT() {
		rep(i, 0, N - 1) mn[i] = 1e9;
	}
	void Upd(int x, int v) {
		for (; x < N; x += x & -x) mn[x] = min(mn[x], v);
	}
	int Qry(int x) {
		int res = 1e9;
		for (; x; x -= x & -x) res = min(res, mn[x]);
		return res;
	}
	void Clear(int x) {
		for (; x < N; x += x & -x) mn[x] = 1e9;
	}
} bit;

void Solve1(int l, int r) {
	if (l == r) return;
	int mid = (l + r) >> 1;
	Solve1(l, mid), Solve1(mid + 1, r);
	vector<Oper> Left, Right;
	rep(i, l, mid) 
		if (b[i].op == 1) Left.push_back(b[i]);
	rep(i, mid + 1, r)
		if (b[i].op == 2) Right.push_back(b[i]);
	int pos = Left.size();
	reo(i, (int)Right.size() - 1, 0) {
		while (pos && Left[pos - 1].x >= Right[i].x) 
			--pos, bit.Upd(N - Left[pos].y, Left[pos].x + Left[pos].y);
		ans[Right[i].id] = min(ans[Right[i].id], bit.Qry(N - Right[i].y) - (Right[i].x + Right[i].y));
	}
	rep(i, pos, (int)Left.size() - 1) bit.Clear(N - Left[i].y);
	int R = l - 1;
	for (int i = l, j = mid + 1; i <= mid || j <= r; )
		if (i <= mid && (j > r || b[i].x <= b[j].x)) c[++R] = b[i++];
		else c[++R] = b[j++];
	rep(i, l, r) b[i] = c[i];
}

void Solve2(int l, int r) {
	if (l == r) return;
	int mid = (l + r) >> 1;
	Solve2(l, mid), Solve2(mid + 1, r);
	vector<Oper> Left, Right;
	rep(i, l, mid)
		if (b[i].op == 1) Left.push_back(b[i]);
	rep(i, mid + 1, r)
		if (b[i].op == 2) Right.push_back(b[i]);
	int pos = Left.size();
	reo(i, (int)Right.size() - 1, 0) {
		while (pos && Left[pos - 1].x >= Right[i].x) 
			--pos, bit.Upd(Left[pos].y, Left[pos].x - Left[pos].y);
		ans[Right[i].id] = min(ans[Right[i].id], bit.Qry(Right[i].y) - (Right[i].x - Right[i].y));
	}
	rep(i, pos, (int)Left.size() - 1) bit.Clear(Left[i].y);
	int R = l - 1;
	for (int i = l, j = mid + 1; i <= mid || j <= r; )
		if (i <= mid && (j > r || b[i].x <= b[j].x)) c[++R] = b[i++];
		else c[++R] = b[j++];
	rep(i, l, r) b[i] = c[i];
}

void Solve3(int l, int r) {
	if (l == r) return;
	int mid = (l + r) >> 1;
	Solve3(l, mid), Solve3(mid + 1, r);
	vector<Oper> Left, Right;
	rep(i, l, mid)
		if (b[i].op == 1) Left.push_back(b[i]);
	rep(i, mid + 1, r)
		if (b[i].op == 2) Right.push_back(b[i]);
	int pos = -1;
	rep(i, 0, (int)Right.size() - 1) {
		while (pos < (int)Left.size() - 1 && Left[pos + 1].x <= Right[i].x)
			++pos, bit.Upd(Left[pos].y, - Left[pos].x - Left[pos].y);
		ans[Right[i].id] = min(ans[Right[i].id], Right[i].x + Right[i].y + bit.Qry(Right[i].y));
	}
	rep(i, 0, pos) bit.Clear(Left[i].y);
	int R = l - 1;
	for (int i = l, j = mid + 1; i <= mid || j <= r; )
		if (i <= mid && (j > r || b[i].x <= b[j].x)) c[++R] = b[i++];
		else c[++R] = b[j++];
	rep(i, l, r) b[i] = c[i];
}

void Solve4(int l, int r) {
	if (l == r) return;
	int mid = (l + r) >> 1;
	Solve4(l, mid), Solve4(mid + 1, r);
	vector<Oper> Left, Right;
	rep(i, l, mid)
		if (b[i].op == 1) Left.push_back(b[i]);
	rep(i, mid + 1, r)
		if (b[i].op == 2) Right.push_back(b[i]);
	int pos = -1;
	rep(i, 0, (int)Right.size() - 1) {
		while (pos < (int)Left.size() - 1 && Left[pos + 1].x <= Right[i].x)
			++pos, bit.Upd(N - Left[pos].y, Left[pos].y - Left[pos].x);
		ans[Right[i].id] = min(ans[Right[i].id], Right[i].x - Right[i].y + bit.Qry(N - Right[i].y));
	}
	rep(i, 0, pos) bit.Clear(N - Left[i].y);
	int R = l - 1;
	for (int i = l, j = mid + 1; i <= mid || j <= r; )
		if (i <= mid && (j > r || b[i].x <= b[j].x)) c[++R] = b[i++];
		else c[++R] = b[j++];
	rep(i, l, r) b[i] = c[i];
}

int main() {
	ios::sync_with_stdio(false), cin.tie(nullptr);
	cin >> n >> m;
	rep(i, 1, n) 
		++tot, a[tot].op = 1, cin >> a[tot].x >> a[tot].y;
	rep(i, 1, m) {
		++tot, cin >> a[tot].op >> a[tot].x >> a[tot].y;
		if (a[tot].op == 2) 
			a[tot].id = ++qtot;
	}
	rep(i, 1, qtot) ans[i] = 2e9;
	rep(i, 1, tot) b[i] = a[i];
	Solve1(1, tot);
	rep(i, 1, tot) b[i] = a[i];
	Solve2(1, tot);
	rep(i, 1, tot) b[i] = a[i];
	Solve3(1, tot);
	rep(i, 1, tot) b[i] = a[i];
	Solve4(1, tot);
	rep(i, 1, qtot) cout << ans[i] << '\n';
	return 0;
}

注意到 K-D Tree 直接秒了。

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平面最近點對 & 最小周長三角形 & 曼哈頓距離最近

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Solution

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Statement

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