BestCoder Round 38-1001 Four Inages Strategy

题目链接：http://bestcoder.hdu.edu.cn/contests/contest_showproblem.php?cid=577&pid=1001

题面：

Four Inages Strategy

Accepts: 272

Submissions: 1374

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

题意：

给定空间四个点，判断四个点是否组成了一个正方形。

解法：

比较笨的考虑了四个点的全部位置分布，不过比较安全。先判断对角线相等， 邻边相等，再加邻边垂直。

代码：

#include <iostream>
#include <string>
#include <cstring>
int store[26],tmp[26];
using namespace std;
int squared_dis(int a,int b,int c,int d,int e,int f)
{
    return (a-d)*(a-d)+(b-e)*(b-e)+(c-f)*(c-f); 
}
int main()
{
    int n,x1,y1,z1,x2,y2,z2,x3,y3,z3,x4,y4,z4;
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        bool flag=false;
        cin>>x1>>y1>>z1>>x2>>y2>>z2>>x3>>y3>>z3>>x4>>y4>>z4;
        if(squared_dis(x1,y1,z1,x3,y3,z3)==squared_dis(x2,y2,z2,x4,y4,z4))
        {
            if(squared_dis(x2,y2,z2,x3,y3,z3)==squared_dis(x1,y1,z1,x2,y2,z2))
            {
                if((x4-x1)*(x2-x1)+(y4-y1)*(y2-y1)+(z4-z1)*(z2-z1)==0)
                flag=true;
            }
        }
        if(squared_dis(x1,y1,z1,x2,y2,z2)==squared_dis(x3,y3,z3,x4,y4,z4))
        {
            if(squared_dis(x2,y2,z2,x3,y3,z3)==squared_dis(x1,y1,z1,x3,y3,z3))
            {
                if((x4-x1)*(x3-x1)+(y4-y1)*(y3-y1)+(z4-z1)*(z3-z1)==0)
                flag=true;
            }
        }
        if(squared_dis(x1,y1,z1,x4,y4,z4)==squared_dis(x2,y2,z2,x3,y3,z3))
        {
            if(squared_dis(x2,y2,z2,x1,y1,z1)==squared_dis(x1,y1,z1,x3,y3,z3))
            {
                if((x3-x1)*(x2-x1)+(y3-y1)*(y2-y1)+(z3-z1)*(z2-z1)==0)
                flag=true;
            }
        }
        cout<<"Case #"<<i<<": ";
        if(flag)cout<<"Yes\n";
        else cout<<"No\n";
    }
    return 0;
}