poj 3335(半平面交)

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 
#include 
#include 
#include 
#include 
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 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& L) {
    vector poly;
    int n = L.size();
    sort(L.begin(), L.end());
    int first = 0, rear = 0; 
    vector p(n); 
    vector 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 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;
}