题目链接：点击打开链接

题意：

给定T表示case数

以下4行是一个case

每行2个点，u v

每次u能够绕着v逆时针转90°

问最少操作多少次使得4个u构成一个正方形。

思路：

枚举判可行

#include <iostream>

#include <cmath>

#include <algorithm>

#include <cstdio>

using namespace std;

int hah,ijj;

int haifei;

template <class T>

inline bool rd(T &ret) {

    char c; int sgn;

    if(c=getchar(),c==EOF) return 0;

    while(c!='-'&&(c<'0'||c>'9')) c=getchar();

    sgn=(c=='-')?

-1:1;

    ret=(c=='-')?

0:(c-'0');

    while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');

    ret*=sgn;

    return 1;

}

template <class T>

inline void pt(T x) {

    if (x <0) {

        putchar('-');

        x = -x;

    }

    if(x>9) pt(x/10);

    putchar(x%10+'0');

}

///////////////////////

const double eps = 1e-8;

const double pi = acos(-1.0);

struct node {

    double x, y;

};

bool dcmp(double i, double j) {

    return fabs(i - j) <= eps;

}

bool eq(const node& i, const node& j) {

    return dcmp(i.x, j.x) && dcmp(i.y, j.y);

}

/*

x0= (x - rx0)*cos(a) - (y - ry0)*sin(a) + rx0 ;

y0= (x - rx0)*sin(a) + (y - ry0)*cos(a) + ry0 ;

*/

node turn(const node& i, const node& j, double a) {

    node re;

    re.x= (i.x - j.x)*cos(a) - (i.y - j.y)*sin(a) + j.x;

    re.y= (i.x - j.x)*sin(a) + (i.y - j.y)*cos(a) + j.y;

    return re;

}

bool cc(const node& i, const node& j) {

    if (!dcmp(i.x, j.x))

        return i.x < j.x;

    else

        return i.y < j.y;

}

double sqr(double x) {

    return x * x;

}

double D(node i, node j) {

    return sqr(i.x-j.x) + sqr(i.y-j.y);

}

double dis[20];

int idx;

bool ok(node i, node j, node k, node z) {

    node ar[4];

    ar[0]=i; ar[1]=j; ar[2]=k; ar[3]=z;

    idx = 0;

    for (int i = 0; i < 4; ++i)

        for (int j = i + 1; j < 4; ++j)

            dis[idx++]=D(ar[i],ar[j]);

    sort(dis, dis +idx);

    if (dcmp(dis[0], dis[3]) && !dcmp(dis[0], 0) && dcmp(dis[4], dis[5]) && dcmp(dis[0] * 2, dis[4])) {

        return true;

    } else

        return false;

}

int main() {

    node a[10], b[10];

    int T;

    rd(T);

    while (T -- > 0) {

        for (int i = 0; i < 4; ++i)

            scanf("%lf%lf%lf%lf", &a[i].x, &a[i].y, &b[i].x, &b[i].y);

        int ans = 100;

        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)

            if (ok(turn(a[0], b[0], i*pi/2),turn(a[1], b[1], j*pi/2),

                   turn(a[2], b[2], k*pi/2),turn(a[3], b[3], l*pi/2))) {

                        ans = min(i+j+k+l, ans);

                   }

        if (ans == 100)

            ans = -1;

        pt(ans);

        putchar('\n');

    }

    return 0;

}