struct point
{double x,y;point(double a = 0,double b = 0): x(a),y(b){}
};

int dblcmp(double x)
{if(fabs(x) < eps)return 0;return x > 0 ? 1:-1;
}
或者
int dblcmp(double x)
{if (x > eps) return 1;else if (x < -eps) return -1;else return 0;
}

double det(double x1, double y1, double x2, double y2)
{    //求叉积return x1*y2 - x2*y1;
}double cross(Point a, Point b, Point c)
{    //向量ab，和向量acreturn det(b.x - a.x, b.y - a.y, c.x - a.x, c.y - a.y);
}double dotdet(double x1, double y1, double x2, double y2)
{    //点积return x1*x2 + y1*y2;
}double dot(Point a, Point b, Point c)
{return dotdet(b.x - a.x, b.y - a.y, c.x - a.x, c.y - a.y);
}int betweenCmp(Point a, Point b, Point c)
{    //判断a是不是在bc范围内return dbcmp(dot(a, b, c));
}bool segcross(Point a, Point b, Point c, Point d)
{double s1, s2;int d1, d2, d3, d4;d1 = dbcmp(s1 = cross(a, b, c));d2 = dbcmp(s2 = cross(a, b, d));d3 = dbcmp(cross(c, d, a));d4 = dbcmp(cross(c, d, b));if((d1^d2) == -2 && (d3^d4) == -2) {    //规范相交return true;}//非规范相交if((d1 == 0 && betweenCmp(c, a, b) <= 0) ||(d2 == 0 && betweenCmp(d, a, b) <= 0) ||(d3 == 0 && betweenCmp(a, c, d) <= 0) ||(d4 == 0 && betweenCmp(b, c, d) <= 0))return true;return false;
}

double area(point p[],int n){ //这里是相对于原点(0, 0)，也可以在多边形上找一个点作为向量的起点double s = 0;int i;p[n].x = p[0].x;p[n].y = p[0].y;for(i = 0; i < n; ++i)s += det(p[i].x, p[i].y, p[i+1].x, p[i+1].y);return fabs(s / 2.0);
}

p = (a+b+c)/2;
S = sqrt(p*(p-a)*(p-b)*(p-c));

A(x1,y1),B(x2,y2),C(x3,y3);
S = fabs(-x2 * y1 + x3*y1+x1*y2-x3*y2-x1*y3+x2*y3) / 2.0;

输入p[1],p[2] ...p[n]。 令p[0] = p[n] p[n + 1] = p[1];　
bool isconvexpg()
{int dir = 0;for (int i = 0; i <= n - 1; ++i){int temp = dblcmp(cross(p[i],p[i + 1],p[i + 2]));if (!dir) dir = temp;if (dir*temp < 0) return false;}return true;
}

double getdis(point a,point b)
{double x = a.x - b.x;double y = a.y - b.y;return sqrt(x*x + y*y);
}
//利用点积求角度
double getangle(point a,point b,point c)
{double dj = dotdet(b.x - a.x,b.y - a.y,c.x - a.x,c.y - a.y);double dis = getdis(a,b)*getdis(a,c);double tmp = dj/dis;return acos(tmp);
}
//判断是否在多边形内
bool IsIn()
{int i;double angle = 0.0;for (i = 1; i <= n; ++i){if (dblcmp(cross(cir,p[i],p[i + 1])) >= 0)angle += getangle(cir,p[i],p[i + 1]);elseangle -= getangle(cir,p[i],p[i + 1]);}//printf("angle == %lf\n",angle);if (dblcmp(angle) == 0) return false;//在外边else if (dblcmp(angle - pi) == 0 || dblcmp(angle + pi) == 0)//在边上{if (dblcmp(r) == 0) return true;}else if (dblcmp(angle - 2.0*pi) == 0 || dblcmp(angle + 2.0*pi) == 0)//在里面{return true;}else //在多边行顶点{if (dblcmp(r) == 0) return true;}return false;
}

//这里起点与终点肯定在凸包上，不知道怎么证明
int graham(point *p,int len)
{top = 0;  int i;//先排序，lrj黑书上的排序方法sort(p,p + len,cmp);stack[top++] = p[0];stack[top++] = p[1];//求右链for (i = 2; i < len; ++i){while (top > 1 && dblcmp(cross(stack[top - 2],stack[top - 1],p[i])) <= 0) top--;stack[top++] = p[i];}//求左链int tmp = top;for (i =len - 2; i >= 0; --i){while (top > tmp && dblcmp(cross(stack[top - 2],stack[top - 1],p[i]))<= 0) top--;stack[top++] = p[i];}return top - 1;//起点两次进栈 - 1
}

int rotaing(point *p,int len)
{p[len] = p[0];int ans = 0,q = 1;for (int i = 0; i < len; ++i){while (cross(p[i],p[i + 1],p[q + 1]) > cross(p[i],p[i + 1],p[q]))q = (q + 1)%len;ans = max(ans,max(dis2(p[i],p[q]),dis2(p[i + 1],p[q + 1])));//这里之所以计算i+1与q+1的距离是考虑到在凸多边形中存在平行边的问题}return ans;
}