(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 =210;
#define clc(a,b) memset(a,b,sizeof(a))
const double eps = 1e-7;
void fre() {freopen("in.txt","r",stdin);}
void freout() {freopen("out.txt","w",stdout);}
inline int read() {int x=0,f=1;char ch=getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch=getchar();}while(ch>='0'&&ch<='9') {x=x*10+ch-'0';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[210];
int main() {
    int T;
    scanf("%d",&T);
    for(int cas=1; cas<=T; cas++) {
        int n,cnt;
        int ans=0,tem=0;
        scanf("%d",&n);
        for(int i=1; i<=n; i++) scanf("%lf%lf%lf",&p[i].x,&p[i].y,&p[i].z);
        for(int i=1; i<n; i++) {
            for(int j=i+1; j<=n; j++) {
                cnt=0;
                for(int k=1; 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=0; k<cnt-1; k++) {
                    for(int h=k+1; 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/=2;
        ans=ans-2*tem/6;
        printf("Case #%d: ",cas);
        printf("%d
",ans);
    }
    return 0;
}