1 CCPC网络赛 HDU5839 Special Tetrahedron
2 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
3 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。
4 要符合条件的话，则该集合里面找两个点，并且要判断一下。
5 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
6
7
8 #include <bits/stdc++.h>
9 using namespace std;
10 #define LL long long
11 const double inf = 123456789012345.0;
12 const LL MOD =100000000LL;
13 const int N =210;
14 #define clc(a,b) memset(a,b,sizeof(a))
15 const double eps = 1e-7;
16 void fre() {freopen("in.txt","r",stdin);}
17 void freout() {freopen("out.txt","w",stdout);}
18 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;}
19
20 struct Point {
21 double x,y,z;
22 Point() {}
23 Point(LL _x,LL _y,LL _z):x(_x),y(_y),z(_z) {}
24 Point operator + (const Point &t) const {
25 return Point(x+t.x,y+t.y,z+t.z);
26 }
27 Point operator -(const Point &t) const {
28 return Point(x-t.x,y-t.y,z-t.z);
29 }
30 Point operator *(const Point &t) const {
31 return Point(y*t.z-z*t.y,z*t.x-x*t.z,x*t.y-y*t.x);
32 }
33 double operator ^(const Point &t) const {
34 return x*t.x+y*t.y+z*t.z;
35 }
36 double len2(){
37 return x*x+y*y+z*z;
38 }
39 } p[N];
40
41 bool check(Point a,Point b,Point c,Point d) {
42 return (((a-b)*(a-c))^(a-d))==0.0;
43 }
44
45 int l[210];
46 int main() {
47 int T;
48 scanf("%d",&T);
49 for(int cas=1; cas<=T; cas++) {
50 int n,cnt;
51 int ans=0,tem=0;
52 scanf("%d",&n);
53 for(int i=1; i<=n; i++) scanf("%lf%lf%lf",&p[i].x,&p[i].y,&p[i].z);
54 for(int i=1; i<n; i++) {
55 for(int j=i+1; j<=n; j++) {
56 cnt=0;
57 for(int k=1; k<=n; k++) {
58 if(k==i||k==j) continue;
59 Point p1,p2;
60 p1=p[k]-p[i];
61 p2=p[k]-p[j];
62 if(p1.len2()==p2.len2()) {
63 l[cnt++]=k;
64 }
65 }
66 for(int k=0; k<cnt-1; k++) {
67 for(int h=k+1; h<cnt; h++) {
68 Point p1=p[l[h]]-p[i],p2=p[l[k]]-p[i];
69 if(p1.len2()!=p2.len2()) continue;
70 if(check(p[i],p[j],p[l[k]],p[l[h]])) continue;
71 ans++;
72 Point p3=p[l[k]]-p[l[h]];
73 Point p4=p[i]-p[j];
74 if(p1.len2()==p3.len2()&&p4.len2()==p3.len2())
75 tem++;
76 }
77 }
78 }
79 }
80 ans/=2;
81 ans=ans-2*tem/6;
82 printf("Case #%d: ",cas);
83 printf("%d\n",ans);
84 }
85 return 0;
86 }
