hdu 5206 判断三维空间中的四个点是否构成正方形

题目描述：如标题

思路：一比一比根号二咯

#include <algorithm>
#include <iostream>
#include <string>
using namespace std;

const int N = 4;

struct Node 
{
    int x, y, z;
} node[N];

inline long long square( int k )
{
    return ( long long ) k * k;
}

long long list[N];

long long dis( Node a, Node b )
{
    return square( a.x - b.x ) + square( a.y - b.y ) + square( a.z - b.z );
}

bool solve()
{
    for ( int i = 0; i < 4; i++ )
    {
        int cnt = 0;
        for ( int j = 0; j < 4; j++ )
        {
            if ( i == j ) continue;
            list[cnt++] = dis( node[i], node[j] );
        }
        sort( list, list + cnt );
        if ( list[0] == list[1] && list[1] * 2 == list[2] ) {}
        else return false;
    }
    return true;
}

int main ()
{
    int t;
    cin >> t;
    for ( int _case = 1; _case <= t; _case++ )
    {
        for ( int i = 0; i < 4; i++ )
        {
            cin >> node[i].x >> node[i].y >> node[i].z;
        }
        cout << "Case #" << _case << ": " << ( solve() ? "Yes" : "No" ) << endl;
    }
    return 0;
}

其实还有许多方法，有时间补上。