枚举两个点当做0号点的相邻两边，判断两边长度是否相等和垂直，然后用向量推最后一个点，比较是否相等

Four Inages Strategy

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description

Young F found a secret record which inherited from ancient times in ancestral home by accident, which named "Four Inages Strategy". He couldn‘t restrain inner exciting, open the record, and read it carefully. " Place four magic stones at four points as array
element in space, if four magic stones form a square, then strategy activates, destroying enemy around". Young F traveled to all corners of the country, and have collected four magic stones finally. He placed four magic stones at four points, but didn‘t know
whether strategy could active successfully. So, could you help him?

Input

Multiple test cases, the first line contains an integer T(no
more than 10000),
indicating the number of cases. Each test case contains twelve integers x1,y1,z1,x2,y2,z2,x3,y3,z3,x4,y4,z4,|x|,|y|,|z|≤100000,representing
coordinate of four points. Any pair of points are distinct.

Output

For each case, the output should occupies exactly one line. The output format is Case #x: ans,
here x is
the data number begins at 1,
if your answer is yes,ans is
Yes, otherwise ans is
No.

Sample Input

2
0 0 0 0 1 0 1 0 0 1 1 0
1 1 1 2 2 2 3 3 3 4 4 4

Sample Output

Case #1: Yes
Case #2: No

Source

BestCoder Round #38 ($)

/* ***********************************************
Author        :CKboss
Created Time  :2015年04月19日 星期日 08时52分45秒
File Name     :A.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

typedef long long int LL;

struct Point
{
	LL x,y,z;
	void input() { cin>>x>>y>>z; }
	LL len() { return x*x+y*y+z*z; }
}p[4];

bool check(int a,int b,int c)
{
	Point pp1 = (Point){p[a].x-p[0].x,p[a].y-p[0].y,p[a].z-p[0].z};
	Point pp2 = (Point){p[b].x-p[0].x,p[b].y-p[0].y,p[b].z-p[0].z};
	Point pp3 = (Point){pp1.x-pp2.x,pp1.y-pp2.y,pp1.z-pp2.z};

	LL Len1=pp1.len();
	LL Len2=pp2.len();
	LL Len3=pp3.len();

	if(Len1!=Len2) return false;
	if(Len1+Len2!=Len3) return false;

	LL dx=p[a].x+p[b].x-p[0].x;
	LL dy=p[a].y+p[b].y-p[0].y;
	LL dz=p[a].z+p[b].z-p[0].z;

	if(dx==p[c].x&&dy==p[c].y&&dz==p[c].z) return true;
	return false;
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

	int T_T,cas=1;
	scanf("%d",&T_T);
	while(T_T--)
	{
		for(int i=0;i<4;i++) p[i].input();
		bool flag=false;
		for(int i=1;i<4;i++)
		{
			if(flag==true) break;
			for(int j=i+1;j<4;j++)
			{
				if(flag==true) break;
				for(int k=1;k<4;k++)
				{
					if(flag==true) break;
					if(k==i||k==j) continue;
					flag=check(i,j,k);
				}
			}
		}
		if(flag==true) printf("Case #%d: Yes\n",cas++);
		else printf("Case #%d: No\n",cas++);
	}

    return 0;
}