hdu 6603 Azshara's deep sea

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6603

题目大意是：给你n个点，g的圆  圆给的是圆心。然后在这n个点求一个凸包，在凸包不相邻的两点并且与圆不相交找到直线。

找到凸包上所有不相交直线的最大数。

这里判断直线和圆的关系使用了面积判断，画个图就知道了。

#include
#define ll long long
using namespace  std;

const double eps = 1e-10;
double dcmp(double x) {
    if(fabs(x) < eps) return 0;
    else return x < 0 ? -1 : 1;
}
 
struct Point {
    double x, y;
    Point(double x=0, double y=0):x(x),y(y) {}
}b[401];

typedef Point Vector;
 
Vector operator - (const Point& A, const Point& B) {
    return Vector(A.x-B.x, A.y-B.y);
}
 
double Cross(const Vector& A, const Vector& B) {
    return A.x*B.y - A.y*B.x;
}

 
double distance(double x1,double y1,double x2,double y2){
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
 
double Dot(const Vector& A, const Vector& B) {
    return A.x*B.x + A.y*B.y;
}
 
bool operator < (const Point& p1, const Point& p2) {
    return p1.x < p2.x || (p1.x == p2.x && p1.y < p2.y);
}
 
bool operator == (const Point& p1, const Point& p2) {
    return p1.x == p2.x && p1.y == p2.y;
}
 
bool SegmentProperIntersection(const Point& a1, const Point& a2, const Point& b1, const Point& b2) {
    double c1 = Cross(a2-a1,b1-a1), c2 = Cross(a2-a1,b2-a1),
           c3 = Cross(b2-b1,a1-b1), c4=Cross(b2-b1,a2-b1);
    return dcmp(c1)*dcmp(c2)<0 && dcmp(c3)*dcmp(c4)<0;
}
 
bool OnSegment(const Point& p, const Point& a1, const Point& a2) {
    return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0;
}
 
// 点集凸包
// 如果不希望在凸包的边上有输入点，把两个 <= 改成 <
// 如果不介意点集被修改，可以改成传递引用
vector ConvexHull(vector p) {
    // 预处理，删除重复点
    sort(p.begin(), p.end());
    p.erase(unique(p.begin(), p.end()), p.end());
 
    int n = p.size();
    int m = 0;
    vector ch(n+1);
    for(int i = 0; i < n; i++) {
        while(m > 1 && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;
        ch[m++] = p[i];
    }
    int k = m;
    for(int i = n-2; i >= 0; i--) {
        while(m > k && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;
        ch[m++] = p[i];
    }
    if(n > 1) m--;
    ch.resize(m);
    return ch;
}
int e[500][500],f[405][405];
int main(){
	int t;
	scanf("%d",&t);
	double  x,y;
	int n,g,r;
	while(t--){
		int  ans=0;
		scanf("%d%d%d",&n,&g,&r);
		vector v;
		for(int i=0;i