Special Tetrahedron

Time Limit: 4000/2000 MS (Java/Others)

Memory Limit: 65536/65536 K (Java/Others)

Problem Description

Given

Input

Intput contains multiple test cases.

The first line is an integer

Output

For each test case,output a line which contains"Case #x: y",x represents the xth test(starting from one),y is the number of Special Tetrahedron.

Sample Input

0 0 0

0 1 1

1 0 1

1 1 0

0 0 0

0 0 2

1 1 1

-1 -1 1

1 -1 1

-1 1 1

1 1 0

1 0 1

0 1 1

Sample Output

Case #1: 1

Case #2: 6

题意：

问最少有四条边相同的四面体有多少个?

思路：

枚举一条边的两点,在枚举另外两点,在枚举第3个点的时候要判断是否是等腰三角形和三点是否共线,在枚举第4个点的时候要判断与前边两点是否是等腰三角形和是否共线(共线可以留在判断是否共面一块判断)，还要判断这两个等腰三角形腰长是否相等;最后来到至少有4条边相等的四面体,如果它是正四面体就被重复算了6次，否则被重复算了两次;所有答案就出来了;

AC代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
//#include <bits/stdc++.h>
#include <stack>
#include <map>
 
using namespace std;
 
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
 
typedef  long long LL;
 
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}
 
const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e5+10;
const int maxn=210;
const double eps=1e-12;

int n;
struct PO
{
    int x,y,z;
}po[maxn];

int check1(PO a,PO b,PO c)//判断共线;
{
    PO temp1,temp2;
    temp1.x=a.x-b.x;temp1.y=a.y-b.y;temp1.z=a.z-b.z;
    temp2.x=c.x-b.x;temp2.y=c.y-b.y;temp2.z=c.z-b.z;
    if(temp1.x*temp2.y==temp1.y*temp2.x&&temp1.x*temp2.z==temp1.z*temp2.x&&temp1.y*temp2.z==temp1.z*temp2.y)return 1;
    return 0;
}
int check2(PO a,PO b,PO c,PO d)//判断共面;
{
    int x1=b.x-a.x,y1=b.y-a.y,z1=b.z-a.z;
    int x2=c.x-a.x,y2=c.y-a.y,z2=c.z-a.z;
    int x3=d.x-a.x,y3=d.y-a.y,z3=d.z-a.z;
    LL t1=(LL)x1*y2*z3+y1*z2*x3+z1*x2*y3;
    LL t2=(LL)z1*x3*y2+x2*y1*z3+x1*z2*y3;
    if(t1==t2)return 1;
    return 0;
}
int dis(PO a,PO b)
{
    return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)+(a.z-b.z)*(a.z-b.z);
}
int solve()
{
    int sum1=0,sum2=0;
    for(int i=1;i<=n;i++)
    {
        for(int j=i+1;j<=n;j++)
        {
            for(int k=1;k<=n;k++)
            {
                if(k==i||k==j)continue;
                int dist=dis(po[k],po[i]);
                if(dist!=dis(po[k],po[j]))continue;
                if(check1(po[i],po[j],po[k]))continue;
                for(int u=k+1;u<=n;u++)
                {
                    if(u==i||u==j)continue;
                    if(dis(po[u],po[i])!=dis(po[u],po[j]))continue;
                    if(dis(po[u],po[i])!=dist)continue;
                   //if(i==1&&j==2&&k==3)cout<<check2(po[k],po[u],po[i],po[j])<<u<<"&%&^^*(
";
                    if(check2(po[u],po[k],po[i],po[j]))continue;
                    if(dis(po[u],po[k])==dist&&dis(po[i],po[j])==dist)sum2++;
                    else sum1++;
                }
            }
        }
    }
    return sum1/2+sum2/6;
}

int main()
{
    int t,Case=0;
    read(t);
    while(t--)
    {
        read(n);
        For(i,1,n)
        {
            read(po[i].x);read(po[i].y);read(po[i].z);
        }
        printf("Case #%d: %d
",++Case,solve());
    }
    return 0;
}