Halfplane Intersection

content/geometry/HalfPlane.h

@@ -0,0 +1,50 @@
+/**
+ * Author: chilli
+ * Date: 2019-05-05
+ * License: CC0
+ * Source: https://github.com/zimpha/algorithmic-library/blob/master/computational-geometry/polygon.cc
+ * Description: Computes the intersection of a set of half-planes. Input is given as a set of planes, facing left.
+ * Output is the convex polygon representing the intersection. The points may have duplicates and be collinear.
+ * Will not fail catastrophically if `eps > sqrt(2)(line intersection error)`. Likely to work for more ranges if
+ * 3 half planes are never guaranteed to intersect at the same point.
+ * Time: O(n \log n)
+ * Status: fuzz-tested, submitted on https://maps19.kattis.com/problems/marshlandrescues
+ * Usage:
+ */
+#pragma once
+
+#include "Point.h"
+#include "sideOf.h"
+#include "lineIntersection.h"
+
+typedef Point<double> P;
+typedef array<P, 2> Line;
+#define sp(a) a[0], a[1]
+#define ang(a) (a[1] - a[0]).angle()
+
+int angDiff(Line a, Line b) { return sgn(ang(a) - ang(b)); }
+bool cmp(Line a, Line b) {
+	int s = angDiff(a, b);
+	return (s ? s : sideOf(sp(a), b[0])) < 0;
+}
+vector<P> halfPlaneIntersection(vector<Line> vs) {
+	const double EPS = sqrt(2) * 1e-8;
+	sort(all(vs), cmp);
+	vector<Line> deq(sz(vs) + 5);
+	vector<P> ans(sz(vs) + 5);
+	deq[0] = vs[0];
+	int ah = 0, at = 0, n = sz(vs);
+	rep(i,1,n+1) {
+		if (i == n) vs.push_back(deq[ah]);
+		if (angDiff(vs[i], vs[i - 1]) == 0) continue;
+		while (ah<at && sideOf(sp(vs[i]), ans[at-1], EPS) < 0)
+			at--;
+		while (i!=n && ah<at && sideOf(sp(vs[i]),ans[ah],EPS)<0)
+			ah++;
+		auto res = lineInter(sp(vs[i]), sp(deq[at]));
+		if (res.first != 1) continue;
+		ans[at++] = res.second, deq[at] = vs[i];
+	}
+	if (at - ah <= 2) return {};
+	return {ans.begin() + ah, ans.begin() + at};
+}

content/geometry/PolygonArea.h

@@ -12,8 +12,11 @@
 #include "Point.h"
 
 template<class T>
-T polygonArea2(vector<Point<T>>& v) {
+T polygonArea2(vector<Point<T>> v) {
+	if (sz(v) < 3) return 0;
 	T a = v.back().cross(v[0]);
-	rep(i,0,sz(v)-1) a += v[i].cross(v[i+1]);
+	rep(i,0,sz(v)-1) {
+		 a += v[i].cross(v[i+1]);
+	}
 	return a;
 }