题意:顺时针给出n个点的凸多边形,问凸多边形的内核。
题解:半平面交模板题,有点坑,用大白上的模板时,判断点是否在有向直线的左边,线上的也算。。
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
const double eps = 1e-9;
const double PI = acos(-1);
double Sqr(double x) { return x * x; }
int dcmp(double x) {
if (fabs(x) < eps)
return 0;
return x > 0 ? 1 : -1;
}
struct Point {
double x, y;
Point(double a = 0, double b = 0): x(a), y(b) {}
};
typedef Point Vector;
typedef vector<Point> Polygon;
Vector operator + (const Vector& a, const Vector& b) { return Vector(a.x + b.x, a.y + b.y); }
Vector operator - (const Vector& a, const Vector& b) { return Vector(a.x - b.x, a.y - b.y); }
Vector operator * (const Vector& a, double b) { return Vector(a.x * b, a.y * b); }
Vector operator / (const Vector& a, double b) { return Vector(a.x / b, a.y / b); }
bool operator == (const Vector& a, const Vector& b) { return !dcmp(a.x - b.x) && !dcmp(a.y - b.y); }
bool operator < (const Vector& a, const Vector& b) { return a.x < b.x || (a.x == b.x && a.y < b.y); }
double Dot(const Vector& a, const Vector& b) { return a.x * b.x + a.y * b.y; }
double Length(const Vector& a) { return sqrt(Dot(a, a)); }
double Cross(const Vector& a, const Vector& b) { return a.x * b.y - a.y * b.x; }
double Angle(const Vector& a, const Vector& b) { return acos(Dot(a, b) / Length(a) / Length(b)); }
Vector Rotate(Vector A, double rad) { return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad)); } //默认逆时针,rad前加负号是顺时针
double angle(Vector v) { return atan2(v.y, v.x); }
struct Line {
Point p;
Vector v;
double ang;
Line() {}
Line(Point a, Vector b): p(a), v(b) { ang = atan2(b.y, b.x); }
bool operator < (const Line& L) const { return ang < L.ang; }
Point point(double a) { return p + v * a; }
};
//点p在有向直线L的左边(线上不算)
bool OnLeft(Line L, Point P) {
return Cross(L.v, P - L.p) >= 0; //就是这里坑了
}
//求两直线的交点,前提交点一定存在
Point GetIntersection(Line a, Line b) {
Vector u = a.p - b.p;
double t = Cross(b.v, u) / Cross(a.v, b.v);
return a.p + a.v * t;
}
//求半面交(用到上面两个公式)
int HalfplaneIntersection(vector<Line>& L) {
vector<Point> poly;
int n = L.size();
sort(L.begin(), L.end());
int first = 0, rear = 0;
vector<Point> p(n);
vector<Line> q(n);
q[first] = L[0];
for (int i = 1; i < n; i++) {
while (first < rear && !OnLeft(L[i], p[rear - 1]))
rear--;
while (first < rear && !OnLeft(L[i], p[first]))
first++;
q[++rear] = L[i];
if (fabs(Cross(q[rear].v, q[rear - 1].v)) < eps) {
rear--;
if (OnLeft(q[rear], L[i].p))
q[rear] = L[i];
}
if (first < rear)
p[rear - 1] = GetIntersection(q[rear - 1], q[rear]);
}
while (first < rear && !OnLeft(q[first], p[rear - 1]))
rear--;
if (rear - first <= 1)
return 0;
p[rear] = GetIntersection(q[rear], q[first]);
for (int i = first; i <= rear; i++)
poly.push_back(p[i]);
return poly.size();
}
Point P[105];
vector<Line> L;
int n;
int main() {
int t;
scanf("%d", &t);
while (t--) {
L.clear();
scanf("%d", &n);
for (int i = 0; i < n; i++)
scanf("%lf%lf", &P[i].x, &P[i].y);
for (int i = 0; i < n; i++)
L.push_back(Line(P[i], P[i] - P[(i + 1) % n]));
int res = HalfplaneIntersection(L);
if (res) printf("YES\n");
else printf("NO\n");
}
return 0;
}