http://acm.hdu.edu.cn/showproblem.php?pid=4273
Rescue
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 337 Accepted Submission(s): 243
Problem Description
I work at NASA outer space rescue team which needs much courage and patient. In daily life, I always receive a lot of mission, and I must complete it right now.
Today, team leader announced me that there is a huge spaceship dropping anchor in the out space, and we should reach there for rescue. As a working principle, at first, we should check whether there are persons living in the spaceship. So we carry a kind of machine called life sensor which can sense the life phenomenon when the distance between the machine and the living is not farther than the sense radius.
I have read the designing paper of the spaceship in advance. It has a form of a convex polyhedron, and we can assume it is isodense. For best control, control center of the whole ship is located at the center of the mass. It is sure that if someone is still alive, he will stay at the control center.
It's unfortunately that I find the door is stocked when I try to enter into the spaceship, so I can only sense the living out of the space ship. Now I have opened the machine and it's time to set the sense radius of it. I wonder the minimal radius of the machine which can allowe me to check whether there are persons living in the spaceship.
Today, team leader announced me that there is a huge spaceship dropping anchor in the out space, and we should reach there for rescue. As a working principle, at first, we should check whether there are persons living in the spaceship. So we carry a kind of machine called life sensor which can sense the life phenomenon when the distance between the machine and the living is not farther than the sense radius.
I have read the designing paper of the spaceship in advance. It has a form of a convex polyhedron, and we can assume it is isodense. For best control, control center of the whole ship is located at the center of the mass. It is sure that if someone is still alive, he will stay at the control center.
It's unfortunately that I find the door is stocked when I try to enter into the spaceship, so I can only sense the living out of the space ship. Now I have opened the machine and it's time to set the sense radius of it. I wonder the minimal radius of the machine which can allowe me to check whether there are persons living in the spaceship.
Input
There are multiple test cases.
The first line contains an integer n indicating the number of vertices of the polyhedron. (4 <= n <= 100)
Each of the next n lines contains three integers xi, yi, zi, the coordinates of the polyhedron vertices (-10,000 <= xi, yi, zi <= 10,000).
It guaranteed that the given points are vertices of the convex polyhedron, and the polyhedron is non-degenerate.
The first line contains an integer n indicating the number of vertices of the polyhedron. (4 <= n <= 100)
Each of the next n lines contains three integers xi, yi, zi, the coordinates of the polyhedron vertices (-10,000 <= xi, yi, zi <= 10,000).
It guaranteed that the given points are vertices of the convex polyhedron, and the polyhedron is non-degenerate.
Output
For each test case, output a float number indicating the minimal radius of the machine. Your answer should accurate up to 0.001.
Sample Input
4 0 0 0 1 0 0 0 1 0 0 0 1 8 0 0 0 0 0 2 0 2 0 0 2 2 2 0 0 2 0 2 2 2 0 2 2 2
Sample Output
0.144 1.000
程序：
#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 109
#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)//向量间求和操作
{
return node(x+p.x,y+p.y,z+p.z);
}
node operator -(const node p)//向量间相减操作
{
return node(x-p.x,y-p.y,z-p.z);
}
node operator *(const node p)//向量间叉乘操作
{
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)//向量乘以一个数
{
return node(x*p,y*p,z*p);
}
node operator /(const double p)//向量除以一个数
{
return node(x/p,y/p,z/p);
}
double operator ^(const node p)//向量间点乘操作
{
return x*p.x+y*p.y+z*p.z;
}
};
struct threeD_convex_hull//三维凸包
{
struct face
{
int a,b,c;
int ok;
};
int n;//初始点数
int cnt;//凸包三角形数
node p[M];//初始点
face f[M*8];//凸包三角形
int to[M][M];//点i到j是属于哪个面
double len(node p)//向量的长度
{
return sqrt(p.x*p.x+p.y*p.y+p.z*p.z);
}
double area(node a,node b,node c)//三个点的面积*2
{
return len((b-a)*(c-a));
}
double volume(node a,node b,node c,node d)//四面体面积*6
{
return (b-a)*(c-a)^(d-a);
}
double ptof(node q,face f)//点与面同向
{
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)//维护凸包，若点q在凸包外则更新凸包
{
f[cur].ok=0;//删除当前面，因为此时它在更大的凸包内部
deal(q,f[cur].b,f[cur].a);
deal(q,f[cur].c,f[cur].b);
deal(q,f[cur].a,f[cur].c);
}
//因为每个三角形的的三边是按照逆时针记录的，所以把边反过来后对应的就是与ab边共线的另一个面
void deal(int q,int a,int b)
{
int fa=to[a][b];//与当前面cnt共边的另一个面
face add;
if(f[fa].ok)//若fa面目前是凸包的表面则继续
{
if(ptof(p[q],f[fa])>eps)//若点q能看到fa面继续深搜fa的三条边，更新新的凸包面
dfs(q,fa);
else//当q点可以看到cnt面的同时看不到a，b共边的fa面，则p和a，b点组成一个新的表面三角形
{
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)//判断两个三角形是否共面
{
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()//构建3D凸包
{
cnt=0;
if(n<4)
return;
int sb=1;
for(int i=1;i<n;i++)//保证前两个点不共点
{
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++)//保证前三个点不共线
{
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++)//保证前四个点不共面
{
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++)//构建初始四面体
{
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++)//倍增法更新凸包
{
for(int j=0;j<cnt;j++)//判断每个点是在当前凸包的内部或者外部
{
if(f[j].ok&&ptof(p[i],f[j])>eps)//若在外部且看到j面继续
{
dfs(i,j);
break;
}
}
}
int tmp=cnt;//把不是凸包上的面删除即ok=0；
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()//凸包表面三角形数目
{
return cnt;
}
int faceCnt()//凸包表面多边形数目
{
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()//凸包的重心
{
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)!=-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();
printf("%.3lf
",hull.min_dis(center));
}
return 0;
}