hdu 3662 3D Convex Hull (求三维凸包面数)模版

There are N points in 3D-space which make up a 3D-Convex hull*. How many faces does the 3D-convexhull have? It is guaranteed that all the points are not in the same plane. 



In case you don’t know the definition of convex hull, here we give you a clarification from Wikipedia: 

*Convex hull: In mathematics, the convex hull, for a set of points X in a real vector space V, is the minimal convex set containing X. 

InputThere are several test cases. In each case the first line contains an integer N indicates the number of 3D-points (3< N <= 300), and then N lines follow, each line contains three numbers x, y, z (between -10000 and 10000) indicate the 3d-position
of a point.
OutputOutput the number of faces of the 3D-Convex hull. 

Sample Input

7
1 1 0
1 -1 0
-1 1 0
-1 -1 0
0 0 1
0 0 0
0 0 -0.1
7
1 1 0
1 -1 0
-1 1 0
-1 -1 0
0 0 1
0 0 0
0 0 0.1

Sample Output

8
5

#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define PR 1e-8
#define N 510
using namespace std;
struct TPoint
{
double x,y,z;
TPoint() {}
TPoint(double _x,double _y,double _z):x(_x),y(_y),z(_z) {}
TPoint operator - (const TPoint p)
{
return TPoint(x-p.x,y-p.y,z-p.z);
}
TPoint operator * (const TPoint p)
{
return TPoint(y*p.z-z*p.y,z*p.x-x*p.z,x*p.y-y*p.x);
}
double operator ^ (const TPoint p)
{
return x*p.x+y*p.y+z*p.z;
}//点积
};
struct fac
{
int a,b,c;//凸包一个面上的三个点的编号
bool ok;//该面是否是最终凸包中的面
};
struct T3dull
{
int n;//初始点数
TPoint ply
;//初始点
int trianglecnt;//凸包上三角形数
fac tri
;//凸包三角形
int vis

;//点i到点j是属于哪个面
double dist(TPoint a)
{
return sqrt(a.x*a.x+a.y*a.y+a.z*a.z);
}//两点长度
double area(TPoint a,TPoint b,TPoint c)
{
return dist((b-a)*(c-a));
}//三角形面积*2
double volume(TPoint a,TPoint b,TPoint c,TPoint d)
{
return (b-a)*(c-a)^(d-a);
}//四面体有向体积*6
double ptoplane(TPoint &p,fac &f)//正：点在面同向
{
TPoint m=ply[f.b]-ply[f.a],n=ply[f.c]-ply[f.a],t=p-ply[f.a];
return (m*n)^t;
}
void deal(int p,int a,int b)
{
int f=vis[a][b];
fac add;
if(tri[f].ok)
{
if((ptoplane(ply[p],tri[f]))>PR) dfs(p,f);
else
{
add.a=b,add.b=a,add.c=p,add.ok=1;
vis[p][b]=vis[a][p]=vis[b][a]=trianglecnt;
tri[trianglecnt++]=add;
}
}
}
void dfs(int p,int cnt)//维护凸包，如果点p在凸包外更新凸包
{
tri[cnt].ok=0;
deal(p,tri[cnt].b,tri[cnt].a);
deal(p,tri[cnt].c,tri[cnt].b);
deal(p,tri[cnt].a,tri[cnt].c);
}
bool same(int s,int e)//判断两个面是否为同一面
{
TPoint a=ply[tri[s].a],b=ply[tri[s].b],c=ply[tri[s].c];
return fabs(volume(a,b,c,ply[tri[e].a]))<PR&&fabs(volume(a,b,c,ply[tri[e].b]))<PR&&fabs(volume(a,b,c,ply[tri[e].c]))<PR;
}
void construct()//构建凸包
{
int i,j;
trianglecnt=0;
if(n<4) return ;
bool tmp=1;
for(int i=1; i<n; i++) //前两点不共点
{
if((dist(ply[0]-ply[i]))>PR)
{
swap(ply[1],ply[i]);
tmp=0;
break;
}
}
if(tmp) return ;
tmp=1;
for(i=2; i<n; i++) //前三点不共线
{
if((dist((ply[0]-ply[1])*(ply[1]-ply[i])))>PR)
{
swap(ply[2],ply[i]);
tmp=0;
break;
}
}
if(tmp) return ;
tmp=1;
for(i=3; i<n; i++) //前四点不共面
{
if(fabs((ply[0]-ply[1])*(ply[1]-ply[2])^(ply[0]-ply[i]))>PR)
{
swap(ply[3],ply[i]);
tmp=0;
break;
}
}
if(tmp) return ;
fac 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((ptoplane(ply[i],add))>0)
swap(add.b,add.c);
vis[add.a][add.b]=vis[add.b][add.c]=vis[add.c][add.a]=trianglecnt;
tri[trianglecnt++]=add;
}
for(int i=4; i<n; i++) //构建更新凸包
{
for(j=0; j<trianglecnt; j++)
{
if(tri[j].ok&&(ptoplane(ply[i],tri[j]))>PR)
{
dfs(i,j);
break;
}
}
}
int cnt=trianglecnt;
trianglecnt=0;
for(i=0; i<cnt; i++)
{
if(tri[i].ok)
tri[trianglecnt++]=tri[i];
}
}
double area()//表面积
{
double ret=0;
for(int i=0; i<trianglecnt; i++)
ret+=area(ply[tri[i].a],ply[tri[i].b],ply[tri[i].c]);
return ret/2.0;
}
double volume()//体积
{
TPoint p(0,0,0);
double ret=0;
for(int i=0; i<trianglecnt; i++)
ret+=volume(p,ply[tri[i].a],ply[tri[i].b],ply[tri[i].c]);
return fabs(ret/6);
}
int facetri()//表面三角形数
{
return trianglecnt;
}
int facepolygon()//表面多边形数
{
int ans=0,i,j,k;
for(i=0; i<trianglecnt; i++)
{
for(j=0,k=1; j<i; j++)
{
if(same(i,j))
{
k=0;
break;
}
}
ans+=k;
}
return ans;
}
} hull;
int main()
{
while(~scanf("%d",&hull.n))
{
int i;
for(i=0; i<hull.n; i++)
scanf("%lf%lf%lf",&hull.ply[i].x,&hull.ply[i].y,&hull.ply[i].z);
hull.construct();
printf("%d\n",hull.facepolygon());
}
return 0;
}