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Hints
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For a fixed cutting line, the area of the smaller cake piece depends only on the line’s distance from the origin. So the strawberry part is not about areas yet; it is about finding the closest valid cutting line to the origin.
A line is valid iff all strawberry points lie in one closed half-plane. If the closest valid line has distance , then the answer is the smaller circular segment area
You only need consider supporting lines of the convex hull of the strawberries. Any valid line can be slid toward the points until it touches the convex hull, making its distance to the origin no larger.
So the geometry reduces to: find the minimum distance from the origin to any edge of the convex hull, where the edge line is considered as a supporting line. Also, if the origin is outside the hull, that minimum distance can be because a line through the origin can keep all points on one side.
Build the convex hull in counterclockwise order. If it has area sign-consistent edges around the origin, the origin is inside/on it and the best distance is the minimum point-to-segment-line distance over hull edges. Otherwise the best distance is . Then plug that distance into the circle-segment formula.
We need cut a circle with one line, all strawberries must stay on one side, and we want the smaller cake piece as big as possible.
The trick is that the cake does not care where the strawberries are directly. For any cutting line, the two areas depend only on the distance from the circle center to that line. Since the center is , call that distance .
The closer the line is to the center, the more balanced the two pieces are. Perfectly through the center gives half the cake. Farther away gives one sad little circular cap. So the whole problem becomes:
Find the minimum possible distance from the origin to a line such that all points lie in one closed half-plane.
After that, the area formula handles the cake.
Area from a line distance
For a circle of radius , a chord whose line is distance from the center cuts off a smaller circular segment of area
This is decreasing as increases. No mystery there: push the knife away from the center and the smaller piece shrinks. So maximizing the smaller piece means minimizing .
What lines can be optimal?
Suppose a valid line has all strawberries on one side, but it does not touch their convex hull. Then there is empty space between the line and the strawberries, so we can slide the line toward them. This keeps all strawberries on the same side and makes the line closer to the origin, or at least no farther.
So an optimal line can be assumed to be a supporting line of the convex hull. In plain English: it kisses the hull without cutting through it. Great, now we only need hull edges, not all possible lines. Infinite geometry problems becoming finite is always the good stuff.
When is the answer half the cake?
If there is a valid line through the origin, then and the smaller piece has area
Such a line exists exactly when all points lie in some closed half-plane whose boundary passes through the origin. That happens when the origin is not strictly inside the convex hull of the points.
Why? If the origin is outside the hull, a separating/supporting line through the origin can put the whole hull on one side. If the origin lies on the hull boundary, also fine. But if the origin is strictly inside the hull, every line through the origin cuts the hull into both sides, so the strawberries cannot all stay on one side.
So:
Computing the distance
Build the convex hull in counterclockwise order. For every hull edge from to , the distance from the origin to the infinite line through that edge is
Since the origin is strictly inside the hull in the only nontrivial case, each hull edge is a supporting line and the perpendicular foot lies on the relevant side; this formula is the distance to that supporting line.
Take the minimum over all edges. That gives .
There are two degenerate hull cases:
How to test whether the origin is strictly inside
For a CCW convex polygon, a point is strictly inside iff it is always to the same side of every directed edge, and never on an edge. For the origin, edge has cross product
For a CCW hull, the origin is strictly inside iff all these values are positive. If any is zero or negative, the origin is outside/on the boundary, and .
Because coordinates are integers and the origin is exact, no floating-point nonsense is needed for this test. Nice.
Algorithm
Complexity is from sorting. The hull scan is linear. Fast enough, no drama.
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void setIO() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
}
struct Point {
ll x, y;
bool operator<(const Point& other) const {
if (x != other.x) return x < other.x;
return y < other.y;
}
bool operator==(const Point& other) const {
return x == other.x && y == other.y;
}
};
ll cross(const Point& a, const Point& b) {
return a.x * b.y - a.y * b.x;
}
ll cross(const Point& a, const Point& b, const Point& c) {
return (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);
}
vector<Point> convexHull(vector<Point> p) {
sort(p.begin(), p.end());
p.erase(unique(p.begin(), p.end()), p.end());
if (p.size() <= 1) return p;
vector<Point> lower, upper;
for (const Point& pt : p) {
while (lower.size() >= 2 && cross(lower[lower.size() - 2], lower.back(), pt) <= 0) {
lower.pop_back();
}
lower.push_back(pt);
}
for (int i = (int)p.size() - 1; i >= 0; --i) {
const Point& pt = p[i];
while (upper.size() >= 2 && cross(upper[upper.size() - 2], upper.back(), pt) <= 0) {
upper.pop_back();
}
upper.push_back(pt);
}
lower.pop_back();
upper.pop_back();
lower.insert(lower.end(), upper.begin(), upper.end());
return lower;
}
int main() {
setIO();
int n;
long double r;
cin >> n >> r;
vector<Point> points(n);
for (auto& p : points) cin >> p.x >> p.y;
vector<Point> hull = convexHull(points);
long double d = 0.0L;
int m = (int)hull.size();
if (m >= 3) {
bool originStrictlyInside = true;
for (int i = 0; i < m; ++i) {
Point a = hull[i];
Point b = hull[(i + 1) % m];
if (cross(a, b) <= 0) {
originStrictlyInside = false;
break;
}
}
if (originStrictlyInside) {
d = r;
for (int i = 0; i < m; ++i) {
Point a = hull[i];
Point b = hull[(i + 1) % m];
long double area2 = fabsl((long double)cross(a, b));
long double dx = (long double)b.x - a.x;
long double dy = (long double)b.y - a.y;
long double len = sqrtl(dx * dx + dy * dy);
d = min(d, area2 / len);
}
}
}
long double ratio = min((long double)1.0, max((long double)0.0, d / r));
long double ans = r * r * acosl(ratio) - d * sqrtl(max((long double)0.0, r * r - d * d));
cout << fixed << setprecision(15) << (double)ans << '\n';
}#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void setIO() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
}
struct Point {
ll x, y;
bool operator<(const Point& other) const {
if (x != other.x) return x < other.x;
return y < other.y;
}
bool operator==(const Point& other) const {
return x == other.x && y == other.y;
}
};
ll cross(const Point& a, const Point& b) {
return a.x * b.y - a.y * b.x;
}
ll cross(const Point& a, const Point& b, const Point& c) {
return (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);
}
vector<Point> convexHull(vector<Point> p) {
sort(p.begin(), p.end());
p.erase(unique(p.begin(), p.end()), p.end());
if (p.size() <= 1) return p;
vector<Point> lower, upper;
for (const Point& pt : p) {
while (lower.size() >= 2 && cross(lower[lower.size() - 2], lower.back(), pt) <= 0) {
lower.pop_back();
}
lower.push_back(pt);
}
for (int i = (int)p.size() - 1; i >= 0; --i) {
const Point& pt = p[i];
while (upper.size() >= 2 && cross(upper[upper.size() - 2], upper.back(), pt) <= 0) {
upper.pop_back();
}
upper.push_back(pt);
}
lower.pop_back();
upper.pop_back();
lower.insert(lower.end(), upper.begin(), upper.end());
return lower;
}
int main() {
setIO();
int n;
long double r;
cin >> n >> r;
vector<Point> points(n);
for (auto& p : points) cin >> p.x >> p.y;
vector<Point> hull = convexHull(points);
long double d = 0.0L;
int m = (int)hull.size();
if (m >= 3) {
bool originStrictlyInside = true;
for (int i = 0; i < m; ++i) {
Point a = hull[i];
Point b = hull[(i + 1) % m];
if (cross(a, b) <= 0) {
originStrictlyInside = false;
break;
}
}
if (originStrictlyInside) {
d = r;
for (int i = 0; i < m; ++i) {
Point a = hull[i];
Point b = hull[(i + 1) % m];
long double area2 = fabsl((long double)cross(a, b));
long double dx = (long double)b.x - a.x;
long double dy = (long double)b.y - a.y;
long double len = sqrtl(dx * dx + dy * dy);
d = min(d, area2 / len);
}
}
}
long double ratio = min((long double)1.0, max((long double)0.0, d / r));
long double ans = r * r * acosl(ratio) - d * sqrtl(max((long double)0.0, r * r - d * d));
cout << fixed << setprecision(15) << (double)ans << '\n';
}