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('
');
}
return 0;
}