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  • 3D Convex Hull HDU 3662 三维凸包

    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<iostream>
    #include<cmath>
    #include<cstring>
    #include<cstdlib>
    #include<algorithm>
    #include <stdio.h>
    using namespace std;
    const int MAXN=1001;
    const int N = 500;
    const double EPS=1e-8;
    int g[MAXN][MAXN];
    struct Point{
    double x,y,z;
    Point(){}
    Point(double xx,double yy,double zz):x(xx),y(yy),z(zz){}
    Point operator -(const Point p1){//两向量之差
    return Point(x-p1.x,y-p1.y,z-p1.z);
    }
    Point operator *(Point p){//叉乘
    return Point(y*p.z-z*p.y,z*p.x-x*p.z,x*p.y-y*p.x);
    }
    double operator ^(Point p){//点乘
    return (x*p.x+y*p.y+z*p.z);
    }
    };
    struct CH3D{
    struct face{
    int a,b,c;//表示凸包一个面上三个点的编号
    bool ok;//表示该面是否属于最终凸包中的面
    };
    int n;//初始顶点数
    Point P[MAXN];//初始顶点
    int num; //凸包表面的三角形数
    face F[8*MAXN];
    //int g[MAXN][MAXN];//凸包表面的三角形
    double vlen(Point a){//向量长度
    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){//叉乘
    return Point((b.y-a.y)*(c.z-a.z)-(b.z-a.z)*(c.y-a.y),-((b.x-a.x)*(c.z-a.z)-(b.z-a.z)*(c.x-a.x)),(b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x));
    }
    double area(Point a,Point b,Point c){//三角形面积*2
    return vlen((b-a)*(c-a));
    }
    double volume(Point a,Point b,Point c,Point d){//四面体有向体积*6
    return (b-a)*(c-a)^(d-a);
    }
    double dblcmp(Point &p,face &f){//正:点在面同向
    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];
    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;
    add.ok=1;
    g[p][b]=g[a][p]=g[b][a]=num;
    F[num++]=add;
    }
    }
    }
    void dfs(int p,int now){
    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;
    }
    void pretreat(){//构建三维凸包
    int i,j,tmp;
    face add;
    bool flag;
    num=0;
    if(n<4) return;
    flag=true;
    for(i=1;i<n;i++){//此段是为了保证前四个点不共面,若以保证,则可去掉
    if(vlen(P[0]-P[i])>EPS){
    swap(P[1],P[i]);
    flag=false; break;
    }
    }
    if(flag) return;
    flag=true;
    for(i=2;i<n;i++){//使前三点不共线
    if(vlen((P[0]-P[1])*(P[1]-P[i]))>EPS){
    swap(P[2],P[i]);
    flag=false; break;
    }
    }
    if(flag) return;
    flag=true;
    for(i=3;i<n;i++){//使前四点不共面
    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];
    }
    }
    int polygon(){//表面多边形个数
    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;
    }
    };
    int main()
    {
    int i,n;
    while(scanf("%d",&n)!=EOF){
    CH3D hull;
    memset(g,0,sizeof(g));
    memset(hull.P,0,sizeof(hull.P));
    hull.n = n;
    for(i=0;i<hull.n;i++){
    scanf("%lf%lf%lf",&hull.P[i].x,&hull.P[i].y,&hull.P[i].z);
    }
    hull.pretreat();
    printf("%d\n",hull.polygon());
    }
    return 0;
    }

     
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  • 原文地址:https://www.cnblogs.com/xxxsans/p/12677197.html
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