题意:在天花板上有3个摄像头,下面有一个凸多面体,问摄像头观测不到的在地面上的面积是多少。
思路:先求出每个摄像头对于凸多面体在xoy平面的投影,然后求凸包,然后利用半平面交来求面积交即可,注意求投影时用到一个结论:
如果在空间有3点共线则满足:(z3-z2)/(z2-z1)=(y3-y2)/(y2-y1)=(x3-x2)/(x2-x1)
代码太乱,有时间再整理。
#include <math.h> #include <stdio.h> #include <iostream> #include <algorithm> using namespace std; const int N=105; const double EPS=1e-8; typedef double DIY; DIY Area; struct Point { DIY x,y; Point() {} Point(DIY _x,DIY _y):x(_x),y(_y){} } p[N]; Point MakeVector(Point &P,Point &Q) { return Point(Q.x-P.x,Q.y-P.y); } DIY CrossProduct(Point P,Point Q) { return P.x*Q.y-P.y*Q.x; } Point MinA; Point stack1[N],stack2[N],stack3[N]; int top1,top2,top3; DIY dist(Point A,Point B) { return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y)); } DIY cross(Point A,Point B,Point C) { return (B.x-A.x)*(C.y-A.y)-(B.y-A.y)*(C.x-A.x); } bool cmp1(Point a,Point b) { DIY k=cross(MinA,a,b); if(k>0) return 1; if(k<0) return 0; return dist(MinA,a)>dist(MinA,b); } void Graham1(Point *p,int n,int &top) { int i; for(i=0; i<n; i++) if(p[i].y<p[0].y||(p[i].y==p[0].y&&p[i].x<p[0].x)) swap(p[i],p[0]); MinA=p[0]; sort(p+1,p+n,cmp1); p[n]=p[0]; stack1[0]=p[0]; stack1[1]=p[1]; stack1[2]=p[2]; top=2; for(i=3; i<=n; i++) { while(cross(stack1[top-1],stack1[top],p[i])<=0&&top>=2) --top; stack1[++top]=p[i]; } } void Graham2(Point *p,int n,int &top) { int i; for(i=0; i<n; i++) if(p[i].y<p[0].y||(p[i].y==p[0].y&&p[i].x<p[0].x)) swap(p[i],p[0]); MinA=p[0]; sort(p+1,p+n,cmp1); p[n]=p[0]; stack2[0]=p[0]; stack2[1]=p[1]; stack2[2]=p[2]; top=2; for(i=3; i<=n; i++) { while(cross(stack2[top-1],stack2[top],p[i])<=0&&top>=2) --top; stack2[++top]=p[i]; } } void Graham3(Point *p,int n,int &top) { int i; for(i=0; i<n; i++) if(p[i].y<p[0].y||(p[i].y==p[0].y&&p[i].x<p[0].x)) swap(p[i],p[0]); MinA=p[0]; sort(p+1,p+n,cmp1); p[n]=p[0]; stack3[0]=p[0]; stack3[1]=p[1]; stack3[2]=p[2]; top=2; for(i=3; i<=n; i++) { while(cross(stack3[top-1],stack3[top],p[i])<=0&&top>=2) --top; stack3[++top]=p[i]; } } DIY MultiCross(Point P,Point Q,Point R) { return CrossProduct(MakeVector(Q,P),MakeVector(Q,R)); } struct halfPlane { Point s,t; DIY angle; halfPlane(){} halfPlane(Point _s,Point _t):s(_s),t(_t){} halfPlane(DIY sx,DIY sy,DIY tx,DIY ty):s(sx,sy),t(tx,ty){} void GetAngle() { angle=atan2(t.y-s.y,t.x-s.x); } } hp[N],q[N]; Point IntersectPoint(halfPlane P,halfPlane Q) { DIY a1=CrossProduct(MakeVector(P.s,Q.t),MakeVector(P.s,Q.s)); DIY a2=CrossProduct(MakeVector(P.t,Q.s),MakeVector(P.t,Q.t)); return Point((P.s.x*a2+P.t.x*a1)/(a2+a1),(P.s.y*a2+P.t.y*a1)/(a2+a1)); } bool cmp2(halfPlane P,halfPlane Q) { if(fabs(P.angle-Q.angle)<EPS) return MultiCross(P.s,P.t,Q.s)>0; return P.angle<Q.angle; } bool IsParallel(halfPlane P,halfPlane Q) { return fabs(CrossProduct(MakeVector(P.s,P.t),MakeVector(Q.s,Q.t)))<EPS; } void HalfPlaneIntersect(int n,int &m) { sort(hp,hp+n,cmp2); int i,l=0,r=1; for(m=i=1; i<n; ++i) if(hp[i].angle-hp[i-1].angle>EPS) hp[m++]=hp[i]; n=m; m=0; q[0]=hp[0];q[1]=hp[1]; for(i=2; i<n; i++) { if(IsParallel(q[r],q[r-1])||IsParallel(q[l],q[l+1])) return; while(l<r&&MultiCross(hp[i].s,hp[i].t,IntersectPoint(q[r],q[r-1]))>0) --r; while(l<r&&MultiCross(hp[i].s,hp[i].t,IntersectPoint(q[l],q[l+1]))>0) ++l; q[++r]=hp[i]; } while(l<r&&MultiCross(q[l].s,q[l].t,IntersectPoint(q[r],q[r-1]))>0) --r; while(l<r&&MultiCross(q[r].s,q[r].t,IntersectPoint(q[l],q[l+1]))>0) ++l; q[++r]=q[l]; for(i=l; i<r; ++i) p[m++]=IntersectPoint(q[i],q[i+1]); } void Solve(Point *p1,Point *p2,Point *p3,int &top1,int &top2,int &top3,int &m) { int i,j; Point a,b; Point O; O.x=O.y=0; int num=0; for(i=0;i<top1;i++) { hp[num]=halfPlane(p1[i],p1[(i+1)%top1]); hp[num].GetAngle(); num++; } for(i=0;i<top2;i++) { hp[num]=halfPlane(p2[i],p2[(i+1)%top2]); hp[num].GetAngle(); num++; } for(i=0;i<top3;i++) { hp[num]=halfPlane(p3[i],p3[(i+1)%top3]); hp[num].GetAngle(); num++; } HalfPlaneIntersect(num,m); Area=0; p[m]=p[0]; for(i=0;i<m;++i) Area+=cross(O,p[i],p[i+1]); if(Area<0) Area=-Area; Area/=2.0; } int main() { int n,m,t,i,j,k; Point p1[N],p2[N],p3[N]; top1=top2=top3=0; Point tmp[N]; DIY x[N],y[N],z[N]; while(cin>>n) { for(i=0; i<n; i++) { cin>>x[i]>>y[i]>>z[i]; } for(i=0;i<3;i++) cin>>tmp[i].x>>tmp[i].y; for(i=0;i<n;i++) { p1[i].y=y[i]+(0-z[i])*(y[i]-tmp[0].y)/(z[i]-100); p1[i].x=x[i]+(0-z[i])*(x[i]-tmp[0].x)/(z[i]-100); } for(i=0;i<n;i++) { p2[i].y=y[i]+(0-z[i])*(y[i]-tmp[1].y)/(z[i]-100); p2[i].x=x[i]+(0-z[i])*(x[i]-tmp[1].x)/(z[i]-100); } for(i=0;i<n;i++) { p3[i].y=y[i]+(0-z[i])*(y[i]-tmp[2].y)/(z[i]-100); p3[i].x=x[i]+(0-z[i])*(x[i]-tmp[2].x)/(z[i]-100); } int k=0; Graham1(p1,n,top1); Graham2(p2,n,top2); Graham3(p3,n,top3); Solve(stack1,stack2,stack3,top1,top2,top3,m); printf("%.2lf\n",Area); } return 0; }
作者:ACdreamers 发表于2013-6-4 21:30:59 原文链接
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