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 <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;}