POJ 3862 Asteroids （三维凸包，求两个凸包重心到表面的最短距离）
Asteroids
| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 297 | Accepted: 65 | Special Judge | ||
Description
Association of Collision Management (ACM) is planning to perform the controlled collision of two asteroids. The asteroids will be slowly brought together and collided at negligible speed. ACM expects asteroids to get attached to each other and form a stable object.
Each asteroid has the form of a convex polyhedron. To increase the chances of success of the experiment ACM wants to bring asteroids together in such manner that their centers of mass are as close as possible. To achieve this, ACM operators can rotate the asteroids and move them independently before bringing them together.
Help ACM to find out what minimal distance between centers of mass can be achieved.
For the purpose of calculating center of mass both asteroids are considered to have constant density.
Each asteroid has the form of a convex polyhedron. To increase the chances of success of the experiment ACM wants to bring asteroids together in such manner that their centers of mass are as close as possible. To achieve this, ACM operators can rotate the asteroids and move them independently before bringing them together.
Help ACM to find out what minimal distance between centers of mass can be achieved.
For the purpose of calculating center of mass both asteroids are considered to have constant density.
Input
Input file contains two descriptions of convex polyhedra.
The first line of each description contains integer number n - the number of vertices of the polyhedron (4 <= n <= 60). The following n lines contain three integer numbers xi, yi, zi each - the coordinates of the polyhedron vertices (-10 4 <= xi, yi, zi <= 10 4). It is guaranteed that the given points are vertices of a convex polyhedron, in particular no point belongs to the convex hull of other points. Each polyhedron is non-degenerate.
The two given polyhedra have no common points.
The first line of each description contains integer number n - the number of vertices of the polyhedron (4 <= n <= 60). The following n lines contain three integer numbers xi, yi, zi each - the coordinates of the polyhedron vertices (-10 4 <= xi, yi, zi <= 10 4). It is guaranteed that the given points are vertices of a convex polyhedron, in particular no point belongs to the convex hull of other points. Each polyhedron is non-degenerate.
The two given polyhedra have no common points.
Output
Output one floating point number - the minimal distance between centers of mass of the asteroids that can be achieved. Your answer must be accurate up to 10
-5.
Sample Input
8 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 5 0 0 5 1 0 6 -1 0 6 0 1 6 0 -1 6
Sample Output
0.75
Source
就是对两个凸包求重心到表面的最短距离。
/* HDU 4273 Rescue 给一个三维凸包，求重心到表面的最短距离 模板题：三维凸包+多边形重心+点面距离 */ #include<stdio.h> #include<algorithm> #include<string.h> #include<math.h> #include<stdlib.h> using namespace std; const int MAXN=550; const double eps=1e-8; struct Point { double x,y,z; Point(){} Point(double xx,double yy,double zz):x(xx),y(yy),z(zz){} //两向量之差 Point operator -(const Point p1) { return Point(x-p1.x,y-p1.y,z-p1.z); } //两向量之和 Point operator +(const Point p1) { return Point(x+p1.x,y+p1.y,z+p1.z); } //叉乘 Point operator *(const Point p) { return Point(y*p.z-z*p.y,z*p.x-x*p.z,x*p.y-y*p.x); } Point operator *(double d) { return Point(x*d,y*d,z*d); } Point operator / (double d) { return Point(x/d,y/d,z/d); } //点乘 double operator ^(Point p) { return (x*p.x+y*p.y+z*p.z); } }; struct CH3D { struct face { //表示凸包一个面上的三个点的编号 int a,b,c; //表示该面是否属于最终凸包上的面 bool ok; }; //初始顶点数 int n; //初始顶点 Point P[MAXN]; //凸包表面的三角形数 int num; //凸包表面的三角形 face F[8*MAXN]; //凸包表面的三角形 int g[MAXN][MAXN]; //向量长度 double vlen(Point a) { return sqrt(a.x*a.x+a.y*a.y+a.z*a.z); } //叉乘 Point cross(const Point &a,const Point &b,const Point &c) { return Point((b.y-a.y)*(c.z-a.z)-(b.z-a.z)*(c.y-a.y), (b.z-a.z)*(c.x-a.x)-(b.x-a.x)*(c.z-a.z), (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x) ); } //三角形面积*2 double area(Point a,Point b,Point c) { return vlen((b-a)*(c-a)); } //四面体有向体积*6 double volume(Point a,Point b,Point c,Point d) { return (b-a)*(c-a)^(d-a); } //正：点在面同向 double dblcmp(Point &p,face &f) { Point m=P[f.b]-P[f.a]; Point n=P[f.c]-P[f.a]; Point t=p-P[f.a]; return (m*n)^t; } void deal(int p,int a,int b) { int f=g[a][b];//搜索与该边相邻的另一个平面 face add; if(F[f].ok) { if(dblcmp(P[p],F[f])>eps) dfs(p,f); else { add.a=b; add.b=a; add.c=p;//这里注意顺序，要成右手系 add.ok=true; g[p][b]=g[a][p]=g[b][a]=num; F[num++]=add; } } } void dfs(int p,int now)//递归搜索所有应该从凸包内删除的面 { F[now].ok=0; deal(p,F[now].b,F[now].a); deal(p,F[now].c,F[now].b); deal(p,F[now].a,F[now].c); } bool same(int s,int t) { Point &a=P[F[s].a]; Point &b=P[F[s].b]; Point &c=P[F[s].c]; return fabs(volume(a,b,c,P[F[t].a]))<eps && fabs(volume(a,b,c,P[F[t].b]))<eps && fabs(volume(a,b,c,P[F[t].c]))<eps; } //构建三维凸包 void create() { int i,j,tmp; face add; num=0; if(n<4)return; //********************************************** //此段是为了保证前四个点不共面 bool flag=true; for(i=1;i<n;i++) { if(vlen(P[0]-P[i])>eps) { swap(P[1],P[i]); flag=false; break; } } if(flag)return; flag=true; //使前三个点不共线 for(i=2;i<n;i++) { if(vlen((P[0]-P[1])*(P[1]-P[i]))>eps) { swap(P[2],P[i]); flag=false; break; } } if(flag)return; flag=true; //使前四个点不共面 for(int i=3;i<n;i++) { if(fabs((P[0]-P[1])*(P[1]-P[2])^(P[0]-P[i]))>eps) { swap(P[3],P[i]); flag=false; break; } } if(flag)return; //***************************************** for(i=0;i<4;i++) { add.a=(i+1)%4; add.b=(i+2)%4; add.c=(i+3)%4; add.ok=true; if(dblcmp(P[i],add)>0)swap(add.b,add.c); g[add.a][add.b]=g[add.b][add.c]=g[add.c][add.a]=num; F[num++]=add; } for(i=4;i<n;i++) { for(j=0;j<num;j++) { if(F[j].ok&&dblcmp(P[i],F[j])>eps) { dfs(i,j); break; } } } tmp=num; for(i=num=0;i<tmp;i++) if(F[i].ok) F[num++]=F[i]; } //表面积 double area() { double res=0; if(n==3) { Point p=cross(P[0],P[1],P[2]); res=vlen(p)/2.0; return res; } for(int i=0;i<num;i++) res+=area(P[F[i].a],P[F[i].b],P[F[i].c]); return res/2.0; } double volume() { double res=0; Point tmp(0,0,0); for(int i=0;i<num;i++) res+=volume(tmp,P[F[i].a],P[F[i].b],P[F[i].c]); return fabs(res/6.0); } //表面三角形个数 int triangle() { return num; } //表面多边形个数 int polygon() { int i,j,res,flag; for(i=res=0;i<num;i++) { flag=1; for(j=0;j<i;j++) if(same(i,j)) { flag=0; break; } res+=flag; } return res; } //三维凸包重心 Point barycenter() { Point ans(0,0,0),o(0,0,0); double all=0; for(int i=0;i<num;i++) { double vol=volume(o,P[F[i].a],P[F[i].b],P[F[i].c]); ans=ans+(o+P[F[i].a]+P[F[i].b]+P[F[i].c])/4.0*vol; all+=vol; } ans=ans/all; return ans; } //点到面的距离 double ptoface(Point p,int i) { return fabs(volume(P[F[i].a],P[F[i].b],P[F[i].c],p)/vlen((P[F[i].b]-P[F[i].a])*(P[F[i].c]-P[F[i].a]))); } }; CH3D hull; int main() { // freopen("in.txt","r",stdin); // freopen("out.txt","w",stdout); while(scanf("%d",&hull.n)==1) { for(int i=0;i<hull.n;i++) { scanf("%lf%lf%lf",&hull.P[i].x,&hull.P[i].y,&hull.P[i].z); } hull.create(); Point p=hull.barycenter(); double ans1=1e20; for(int i=0;i<hull.num;i++) { ans1=min(ans1,hull.ptoface(p,i)); } scanf("%d",&hull.n); for(int i=0;i<hull.n;i++) { scanf("%lf%lf%lf",&hull.P[i].x,&hull.P[i].y,&hull.P[i].z); } hull.create(); p=hull.barycenter(); double ans2=1e20; for(int i=0;i<hull.num;i++) { ans2=min(ans2,hull.ptoface(p,i)); } printf("%.5f\n",ans1+ans2); } return 0; }