Asteroids
| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 481 | Accepted: 152 | 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 (-104 <= xi, yi, zi <= 104). 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 (-104 <= xi, yi, zi <= 104). 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
分析：分别求出重心到面的最短距离：
#include"stdio.h" #include"string.h" #include"iostream" #include"map" #include"string" #include"queue" #include"stack" #include"vector" #include"stdlib.h" #include"algorithm" #include"math.h" #define M 533 #define eps 1e-10 #define inf 0x3f3f3f3f #define mod 1070000009 #define PI acos(-1.0) using namespace std; struct node { double x,y,z,dis; node(){} node(double xx,double yy,double zz):x(xx),y(yy),z(zz){} node operator +(const node p) { return node(x+p.x,y+p.y,z+p.z); } node operator -(const node p) { return node(x-p.x,y-p.y,z-p.z); } node operator *(const node p) { return node(y*p.z-z*p.y,z*p.x-x*p.z,x*p.y-y*p.x); } node operator *(const double p) { return node(x*p,y*p,z*p); } node operator /(const double p) { return node(x/p,y/p,z/p); } double operator ^(const node p) { return x*p.x+y*p.y+z*p.z; } }; struct threeD_convex_hull { struct face { int a,b,c; int ok; }; int n; int cnt; node p[M]; face f[M*8]; int to[M][M]; double len(node p) { return sqrt(p.x*p.x+p.y*p.y+p.z*p.z); } double area(node a,node b,node c) { return len((b-a)*(c-a)); } double volume(node a,node b,node c,node d) { return (b-a)*(c-a)^(d-a); } double ptof(node q,face f) { node m=p[f.b]-p[f.a]; node n=p[f.c]-p[f.a]; node t=q-p[f.a]; return m*n^t; } void dfs(int q,int cur) { f[cur].ok=0; deal(q,f[cur].b,f[cur].a); deal(q,f[cur].c,f[cur].b); deal(q,f[cur].a,f[cur].c); } void deal(int q,int a,int b) { int fa=to[a][b]; face add; if(f[fa].ok) { if(ptof(p[q],f[fa])>eps) dfs(q,fa); else { add.a=b; add.b=a; add.c=q; add.ok=1; to[b][a]=to[a][q]=to[q][b]=cnt; f[cnt++]=add; } } } int same(int s,int t) { node a=p[f[s].a]; node b=p[f[s].b]; node c=p[f[s].c]; if(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) return 1; return 0; } void make() { cnt=0; if(n<4) return; int sb=1; for(int i=1;i<n;i++) { if(len(p[0]-p[i])>eps) { swap(p[1],p[i]); sb=0; break; } } if(sb)return; sb=1; for(int i=2;i<n;i++) { if(len((p[1]-p[0])*(p[i]-p[0]))>eps) { swap(p[2],p[i]); sb=0; break; } } if(sb)return; sb=1; for(int i=3;i<n;i++) { if(fabs(volume(p[0],p[1],p[2],p[i]))>eps) { swap(p[3],p[i]); sb=0; break; } } if(sb)return; face add; for(int i=0;i<4;i++) { add.a=(i+1)%4; add.b=(i+2)%4; add.c=(i+3)%4; add.ok=1; if(ptof(p[i],add)>eps) swap(add.c,add.b); to[add.a][add.b]=to[add.b][add.c]=to[add.c][add.a]=cnt; f[cnt++]=add; } for(int i=4;i<n;i++) { for(int j=0;j<cnt;j++) { if(f[j].ok&&ptof(p[i],f[j])>eps) { dfs(i,j); break; } } } int tmp=cnt; cnt=0; for(int i=0;i<tmp;i++) if(f[i].ok) f[cnt++]=f[i]; } double Area()//表面积 { double S=0; if(n==3) { S=area(p[0],p[1],p[2])/2.0; return S; } for(int i=0;i<cnt;i++) S+=area(p[f[i].a],p[f[i].b],p[f[i].c]); return S/2.0; } double Volume()//体积 { double V=0; node mid(0,0,0); for(int i=0;i<cnt;i++) V+=volume(p[f[i].a],p[f[i].b],p[f[i].c],mid); V=fabs(V)/6.0; return V; } int tringleCnt() { return cnt; } int faceCnt() { int num=0; for(int i=0;i<cnt;i++) { int flag=1; for(int j=0;j<i;j++) { if(same(i,j)) { flag=0; break; } } num+=flag; } return num; } double pf_dis(face f,node q)//点到面的距离 { double V=volume(p[f.a],p[f.b],p[f.c],q); double S=area(p[f.a],p[f.b],p[f.c]); return fabs(V/S); } double min_dis(node q)//暴力搜索内部的点q到面的最短距离即体积/面积 { double mini=inf; for(int i=0;i<cnt;i++) { double h=pf_dis(f[i],q); if(mini>h) mini=h; } return mini; } node barycenter() { node ret(0,0,0),mid(0,0,0); double sum=0; for(int i=0;i<cnt;i++) { double V=volume(p[f[i].a],p[f[i].b],p[f[i].c],mid); ret=ret+(mid+p[f[i].a]+p[f[i].b]+p[f[i].c])/4.0*V; sum+=V; } ret=ret/sum; return ret; } }hull; /*int main() { while(scanf("%d",&hull.n)!=EOF) { 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.make(); printf("%d ",hull.faceCnt()); } return 0; }*/ int main() { 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.make(); node center=hull.barycenter(); double min1=hull.min_dis(center); 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.make(); center=hull.barycenter(); double min2=hull.min_dis(center); printf("%.5lf ",min1+min2); } return 0; }