uvalive 2512

题意：一个n边形的房间，在里面放一个监控，能看到整个房间，问有多大面积的区域可以放这个监控。
题解：直接半平面交并计算面积就可以了。

#include <cstdio>#include <cstring>#include <vector>#include <cmath>#include <algorithm>using namespace std;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();}const int N = 1505;int n;Point P[N];vector<Line> L;int main() {    int t;    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;}