HDU 4273 Rescue（三维凸包 + 重心）

Rescue

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 398 Accepted Submission(s): 296



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.



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.



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

Source

2012 ACM/ICPC Asia Regional Changchun Online

kuangbin大神的模板一遍过。。。膜拜一下

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#define eps 1e-8
#define N 110
using namespace std;
struct point
{
    double x,y,z;
    point() {}
    point(double xx,double yy,double zz): x(xx),y(yy),z(zz) {}
    point operator -(const point p)
    {
        return point(x-p.x, y-p.y, z-p.z);
    }
    point operator +(const point p)
    {
        return point(x+p.x, y+p.y, z+p.z);
    }
    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 p)
    {
        return point(x*p, y*p,z*p);
    }
    point operator /(double p)
    {
        return point(x/p, y/p,z/p);
    }
    double operator ^( point p)
    {
        return x*p.x+y*p.y+z*p.z;
    }

};
struct node
{
    struct face
    {
        int a,b,c;
        bool ok;
    };
    int n;
    point tn
;
    int num;
    face F[8*N];
    int g

;
    double vlen(point a)
    {
        return sqrt(a.x*a.x+a.y*a.y+a.z*a.z);
    }
    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)

                     );
    }
    double area(point a,point b,point c)
    {
        return vlen((b*a)*(c-a));
    }
    double volume(point a,point b,point c,point d)
    {
        return (b-a)*(c-a)^(d-a);
    }
    double dbcmp(point &p,face &f)
    {
        point m = tn[f.b] - tn[f.a];
        point n = tn[f.c] - tn[f.a];
        point t = p - tn[f.a];
        return (m*n)^t;
    }
    void deal(int p,int a,int b)
    {
        int f = g[a][b];
        face add;
        if(F[f].ok)
        {
            if(dbcmp(tn[p],F[f]) > eps)
                dfs(p,f);
            else
            {
                add.a=b;
                add.b=a;
                add.c=p;
                add.ok=true;
                g[p][b] = g[a][p] = g[b][a] =num;
                F[num++] = add;
            }
        }
    }
    void dfs(int p,int now)
    {
        F[now].ok = false;
        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 = tn[F[s].a];
        point &b = tn[F[s].b];
        point &c = tn[F[s].c];
        return fabs( volume(a,b,c,tn[F[t].a])) < eps &&
               fabs( volume(a,b,c,tn[F[t].b])) < eps &&
               fabs( volume(a,b,c,tn[F[t].c])) < eps ;
    }
    void create()
    {
        int i,j,tmp;
        face add;
        num = 0;
        if(n<4) return ;
        bool flag =true;
        for(i=1; i<n; i++)
        {
            if(vlen(tn[0]-tn[i]) > eps)
            {
                swap(tn[1],tn[i]);
                flag = false;
                break;
            }
        }
        if(flag) return ;
        flag =true;
        for(i=2; i<n; i++)
        {
            if(vlen((tn[0]-tn[1])*(tn[1]-tn[i])) > eps )
            {
                swap(tn[2],tn[i]);
                flag =false;
                break;
            }
        }
        if(flag) return ;
        flag =true;
        for(i =3; i<n; i++)
        {
            if(fabs((tn[0]-tn[1])*(tn[1]-tn[2])^(tn[0]-tn[i])) >eps    )
            {
                swap(tn[3],tn[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(dbcmp(tn[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 && dbcmp(tn[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];
        }
    }
    double area()
    {
        double res=0;
        if(n==3)
        {
            point p = cross(tn[0],tn[1],tn[2]);
            res = vlen(p)/2.0;
            return res;
        }
        for(int i=0; i<num; i++)
        {
            res += area(tn[F[i].a],tn[F[i].b],tn[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,tn[F[i].a],tn[F[i].b],tn[F[i].c]);
        return fabs(res/6.0);
    }
    int triangle()
    {
        return num;
    }
    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;
    }
    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,tn[F[i].a],tn[F[i].b],tn[F[i].c]);
            ans = ans+(o+tn[F[i].a]+tn[F[i].b]+tn[F[i].c])/4.0*vol;
            all+=vol;
        }
        ans = ans/all;
        return ans;
    }
    double ptoface(point p,int i)
    {
        return fabs(volume(tn[F[i].a],tn[F[i].b],tn[F[i].c],p) /
                    vlen( (tn[F[i].b]-tn[F[i].a])*(tn[F[i].c]-tn[F[i].a]) ));
    }
};
node ans;
int main()
{
    while(~scanf("%d",&ans.n))
    {
        for(int i=0; i<ans.n; i++)
        {
            scanf("%lf%lf%lf",&ans.tn[i].x,&ans.tn[i].y,&ans.tn[i].z);
        }
        ans.create();
        point p=ans.barycenter();
        double answer=1e20;
        for(int i=0; i<ans.num; i++)
        {
            answer=min(answer,ans.ptoface(p,i));
        }
        printf("%.3lf\n",answer);
    }
    return 0;
}