(四面体)CCPC网络赛 HDU5839 Special Tetrahedron

 CCPC网络赛 HDU5839 Special Tetrahedron
 题意：n个点，选四个出来组成四面体，要符合四面体至少四条边相等，若四条边相等则剩下两条边不相邻，求个数
 思路：枚举四面体上一条线，再找到该线两个端点相等的点，放在一个集合里面。
 要符合条件的话，则该集合里面找两个点，并且要判断一下。
 注意，普通四面体会被重复计算两次，正四面体会重复计算六次

 #include <bits/stdc++.h>
 using namespace std;
 #define LL long long
 const double inf = 123456789012345.0;
 const LL MOD =100000000LL;
 const int N =;
 #define clc(a,b) memset(a,b,sizeof(a))
 const double eps = 1e-;
 void fre() {freopen("in.txt","r",stdin);}
 void freout() {freopen("out.txt","w",stdout);}
 inline int read() {int x=,f=;char ch=getchar();while(ch>''||ch<'') {if(ch=='-') f=-; ch=getchar();}while(ch>=''&&ch<='') {x=x*+ch-'';ch=getchar();}return x*f;}

 struct Point {
     double x,y,z;
     Point() {}
     Point(LL _x,LL _y,LL _z):x(_x),y(_y),z(_z) {}
     Point operator + (const Point &t) const {
         return Point(x+t.x,y+t.y,z+t.z);
     }
     Point operator -(const Point &t) const {
         return Point(x-t.x,y-t.y,z-t.z);
     }
     Point operator *(const Point &t) const {
         return Point(y*t.z-z*t.y,z*t.x-x*t.z,x*t.y-y*t.x);
     }
     double operator ^(const Point &t) const {
         return x*t.x+y*t.y+z*t.z;
     }
     double len2(){
         return x*x+y*y+z*z;
     }
 } p[N];

 bool check(Point a,Point b,Point c,Point d) {
     return (((a-b)*(a-c))^(a-d))==0.0;
 }

 int l[];
 int main() {
     int T;
     scanf("%d",&T);
     for(int cas=; cas<=T; cas++) {
         int n,cnt;
         int ans=,tem=;
         scanf("%d",&n);
         for(int i=; i<=n; i++) scanf("%lf%lf%lf",&p[i].x,&p[i].y,&p[i].z);
         for(int i=; i<n; i++) {
             for(int j=i+; j<=n; j++) {
                 cnt=;
                 for(int k=; k<=n; k++) {
                     if(k==i||k==j) continue;
                     Point p1,p2;
                     p1=p[k]-p[i];
                     p2=p[k]-p[j];
                     if(p1.len2()==p2.len2()) {
                         l[cnt++]=k;
                     }
                 }
                 for(int k=; k<cnt-; k++) {
                     for(int h=k+; h<cnt; h++) {
                         Point p1=p[l[h]]-p[i],p2=p[l[k]]-p[i];
                         if(p1.len2()!=p2.len2()) continue;
                         if(check(p[i],p[j],p[l[k]],p[l[h]])) continue;
                         ans++;
                         Point p3=p[l[k]]-p[l[h]];
                         Point p4=p[i]-p[j];
                         if(p1.len2()==p3.len2()&&p4.len2()==p3.len2())
                             tem++;
                     }
                 }
             }
         }
         ans/=;
         ans=ans-*tem/;
         printf("Case #%d: ",cas);
         printf("%d\n",ans);
     }
     return ;
 }