C. Captain Marmot （Codeforces Round #271)

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本文介绍了一种通过旋转鼹鼠位置以形成正方形编队的算法。该算法旨在解决军事演习中鼹鼠编队的问题，通过计算最少的旋转次数使四个鼹鼠形成一个紧凑的正方形。

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Captain Marmot wants to prepare a huge and important battle against his enemy, Captain Snake. For this battle he has n regiments, each consisting of 4 moles.

Initially, each mole i (1 ≤ i ≤ 4n) is placed at some position (xi, yi) in the Cartesian plane. Captain Marmot wants to move some moles to make the regiments compact, if it's possible.

Each mole i has a home placed at the position (ai, bi). Moving this mole one time means rotating his position point (xi, yi) 90 degrees counter-clockwise around it's home point (ai, bi).

A regiment is compact only if the position points of the 4 moles form a square with non-zero area.

Help Captain Marmot to find out for each regiment the minimal number of moves required to make that regiment compact, if it's possible.

Input

The first line contains one integer n (1 ≤ n ≤ 100), the number of regiments.

The next 4n lines contain 4 integers xi, yi, ai, bi ( - 104 ≤ xi, yi, ai, bi ≤ 104).

Output

Print n lines to the standard output. If the regiment i can be made compact, the i-th line should contain one integer, the minimal number of required moves. Otherwise, on the i-th line print "-1" (without quotes).

Sample test(s)

input
4
1 1 0 0
-1 1 0 0
-1 1 0 0
1 -1 0 0
1 1 0 0
-2 1 0 0
-1 1 0 0
1 -1 0 0
1 1 0 0
-1 1 0 0
-1 1 0 0
-1 1 0 0
2 2 0 1
-1 0 0 -2
3 0 0 -2
-1 1 -2 0

output
1
-1
3
3

Note

In the first regiment we can move once the second or the third mole.

We can't make the second regiment compact.

In the third regiment, from the last 3 moles we can move once one and twice another one.

In the fourth regiment, we can move twice the first mole and once the third mole.

四个点绕给定的中心点逆时针旋转90度，看是否能构成正方形，若能，输出最小旋转次数。

每个点最多旋转3次，4次则回到了原来的位置。就可以用个二维数组把每个状态记录下来，最后再判断一下是

否能构成正方形就行了。

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
const int inf=9999999;
using namespace std;
struct node
{
    int x;
    int y;
}p[5][5],home[5];
long long d[8];
long long dis(node a,node b)//距离的平方
{
    return (long long)(a.x-b.x)*(a.x-b.x)+(long long)(a.y-b.y)*(a.y-b.y);
}
void solve()
{
    int ans=inf;
    for(int i=0;i<4;i++)
    {
        for(int j=0;j<4;j++)
        {
            for(int k=0;k<4;k++)
            {
                for(int l=0;l<4;l++)
                {
                    d[0]=dis(p[i][0],p[j][1]);//四边距离的平方
                    d[1]=dis(p[j][1],p[k][2]);
                    d[2]=dis(p[k][2],p[l][3]);
                    d[3]=dis(p[l][3],p[i][0]);
                    d[4]=dis(p[i][0],p[k][2]);//对角线的平方
                    d[5]=dis(p[j][1],p[l][3]);

                    sort(d,d+6);
                    if(d[0]==0)
                    continue;
                    else if(d[0]==d[1]&&d[1]==d[2]&&d[2]==d[3]&&2*d[3]==d[4]&&d[4]==d[5])//判断是否为正方形
                    {
                        ans=min(ans,i+j+k+l);
                    }
                }
            }
        }
    }
    if(ans!=inf)
    printf("%d\n",ans);
    else
    printf("-1\n");
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        for(int i=0;i<4;i++)
        {
            scanf("%d%d",&p[0][i].x,&p[0][i].y);
            scanf("%d%d",&home[i].x,&home[i].y);
            int x=p[0][i].x-home[i].x;
            int y=p[0][i].y-home[i].y;

            p[1][i].x=home[i].x-y;//逆时针旋转90度
            p[1][i].y=home[i].y+x;

            p[2][i].x=home[i].x-x;
            p[2][i].y=home[i].y-y;

            p[3][i].x=home[i].x+y;
            p[3][i].y=home[i].y-x;
        }
        solve();
    }
    return 0;
}