LA 2512 半平面交

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1500 + 10;
const double eps = 1e-9;
const double PI = acos(-1);

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)); }

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; }
};

double PolygonArea(vector<Point>& res, int m)
{
    double area = 0;
    for (int i = 1; i < m - 1; i++)
        area += Cross(res[i] - res[0], res[i + 1] - res[0]);
    return area / 2;
}
//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;
}
//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
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();
}
int n, T;
Point P[maxn];
vector<Line>L;
int main(int argc, char const *argv[])
{
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d", &n);
        for (int i = n - 1; i >= 0; i--)
            scanf("%lf%lf", &P[i].x, &P[i].y);
        L.clear();
        for (int i = 0; i < n; i++)
            L.push_back(Line(P[i], P[(i + 1) % n] - P[i]));
        Polygon poly;
        int cnt = HalfplaneIntersection(L, poly);
        printf("%.2lf\n", PolygonArea(poly, cnt));
    }
    return 0;
}

纯净的半平面交模板题。