/*poj 1279 Art Gallery - 求多边形核的面积*/#include<stdio.h>#include<math.h>#include <algorithm> using namespace std; const double eps=1e-8;struct point {double x,y;}dian[20000+10];point jiao[203];struct line { point s,e; double angle; }xian[20000+10]; int n,yong;bool mo_ee(double x,double y) { double ret=x-y; if(ret<0) ret=-ret; if(ret<eps) return 1; return 0; } bool mo_gg(double x,double y) { return x > y + eps;} // x > y bool mo_ll(double x,double y) { return x < y - eps;} // x < y bool mo_ge(double x,double y) { return x > y - eps;} // x >= y bool mo_le(double x,double y) { return x < y + eps;} // x <= y point mo_intersection(point u1,point u2,point v1,point v2) { point ret=u1; double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x)) /((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x)); ret.x+=(u2.x-u1.x)*t; ret.y+=(u2.y-u1.y)*t; return ret; } double mo_xmult(point p2,point p0,point p1)//p1在p2左返回负，在右边返回正 { return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y); } void mo_HPI_addl(point a,point b) { xian[yong].s=a; xian[yong].e=b; xian[yong].angle=atan2(b.y-a.y,b.x-a.x); yong++; } //半平面交bool mo_HPI_cmp(const line& a,const line& b){if(mo_ee(a.angle,b.angle)){return mo_gg( mo_xmult(b.e,a.s,b.s),0);}else{return mo_ll(a.angle,b.angle);}}int mo_HPI_dq[20000+10];bool mo_HPI_isout(line cur,line top,line top_1){point jiao=mo_intersection(top.s,top.e,top_1.s,top_1.e);return mo_ll( mo_xmult(cur.e,jiao,cur.s),0);}int mo_HalfPlaneIntersect(line *xian,int n,point *jiao){int i,j,ret=0;sort(xian,xian+n,mo_HPI_cmp);for (i = 0, j = 0; i < n; i++){if (mo_gg(xian[i].angle,xian[j].angle)){xian[++j] = xian[i];}}n=j+1;mo_HPI_dq[0]=0;mo_HPI_dq[1]=1;int top=1,bot=0;for (i = 2; i < n; i++){ while (top > bot && mo_HPI_isout(xian[i], xian[mo_HPI_dq[top]], xian[mo_HPI_dq[top-1]])) top--; while (top > bot && mo_HPI_isout(xian[i], xian[mo_HPI_dq[bot]], xian[mo_HPI_dq[bot+1]])) bot++; mo_HPI_dq[++top] = i; //当前半平面入栈} while (top > bot && mo_HPI_isout(xian[mo_HPI_dq[bot]], xian[mo_HPI_dq[top]], xian[mo_HPI_dq[top-1]])) top--; while (top > bot && mo_HPI_isout(xian[mo_HPI_dq[top]], xian[mo_HPI_dq[bot]], xian[mo_HPI_dq[bot+1]])) bot++; mo_HPI_dq[++top] = mo_HPI_dq[bot]; for (ret = 0, i = bot; i < top; i++, ret++){jiao[ret]=mo_intersection(xian[mo_HPI_dq[i+1]].s,xian[mo_HPI_dq[i+1]].e,xian[mo_HPI_dq[i]].s,xian[mo_HPI_dq[i]].e);}return ret;}//求多边形面积double mo_area_polygon(point *dian,int n){int i;point yuan;yuan.x=yuan.y=0;double ret=0;for(i=0;i<n;++i){ret+=mo_xmult(dian[(i+1)%n],yuan,dian[i]);}return ret;}int main(){int i,iofcase=1,t;scanf("%d",&t);while(t--){scanf("%d",&n);yong=0;for(i=0;i<n;++i){scanf("%lf%lf",&dian[i].x,&dian[i].y);}double area=mo_area_polygon(dian,n);if(area<0)//若是顺时针{for(i=0;i<n;++i){mo_HPI_addl(dian[(i+1)%n],dian[i]);}}else{for(i=0;i<n;++i){mo_HPI_addl(dian[i],dian[(i+1)%n]);}}int ret=mo_HalfPlaneIntersect(xian,n,jiao);area=mo_area_polygon(jiao,ret);if(area<0) area=-area;area=area/2;printf("%.2lf\n",area);}return 0;}
