4*4*4*4暴力+点的旋转+推断正方型

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

1
-1
3
3

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>using namespace std;const double pi=3.1415926/2.;
const int INF=0x3f3f3f3f;int n;
struct PT
{double x,y,hx,hy;
}pt[10][10];struct Point
{int x,y;
}A,B,C,D;int isSQRT(Point a,Point b,Point c,Point d)
{int i,j,sumt=0,sumo=0;double px[10],py[10];//代表6条边的 向量double y[10];memset(y,0,sizeof(y));px[1]=a.x-b.x;py[1]=a.y-b.y;px[2]=a.x-c.x;py[2]=a.y-c.y;px[3]=a.x-d.x;py[3]=a.y-d.y;px[4]=b.x-c.x;py[4]=b.y-c.y;px[5]=b.x-d.x;py[5]=b.y-d.y;px[6]=c.x-d.x;py[6]=c.y-d.y;for(i=1; i<=6; i++){for(j=i+1; j<=6; j++)if((px[i]*px[j]+py[i]*py[j])==0)//推断垂直{y[i]++;y[j]++;}}for(i=1; i<=6; i++){if(y[i]==2)sumt++;//有2条边 与其垂直的个数if(y[i]==1)sumo++;//有1条边 与其垂直的个数}if(sumt==4&&sumo==2)return 1;// 是正方形if(sumt==4)return 0;//是 矩形return 0;//都不是
}int main()
{scanf("%d",&n);while(n--){for(int i=0;i<4;i++){double a,b,c,d;scanf("%lf%lf%lf%lf",&a,&b,&c,&d);pt[i][0].x=a,pt[i][0].y=b;pt[i][0].hx=c,pt[i][0].hy=d;///....for(int j=1;j<4;j++){pt[i][j].x=pt[i][j-1].x;pt[i][j].y=pt[i][j-1].y;pt[i][j].hx=pt[i][j-1].hx;pt[i][j].hy=pt[i][j-1].hy;///x0= (x - rx0)*cos(a) - (y - ry0)*sin(a)  + rx0 ;///y0= (x - rx0)*sin(a) + (y - ry0)*cos(a) + ry0 ;pt[i][j].x=(pt[i][j-1].x-pt[i][j-1].hx)*0-(pt[i][j-1].y-pt[i][j-1].hy)*1+pt[i][j-1].hx;pt[i][j].y=(pt[i][j-1].x-pt[i][j-1].hx)*1+(pt[i][j-1].y-pt[i][j-1].hy)*0+pt[i][j-1].hy;}}int ans=INF;for(int i1=0;i1<4;i1++){for(int i2=0;i2<4;i2++){for(int i3=0;i3<4;i3++){for(int i4=0;i4<4;i4++){int temp=i1+i2+i3+i4;if(temp>ans) continue;A.x=pt[0][i1].x;A.y=pt[0][i1].y;B.x=pt[1][i2].x;B.y=pt[1][i2].y;C.x=pt[2][i3].x;C.y=pt[2][i3].y;D.x=pt[3][i4].x;D.y=pt[3][i4].y;if(isSQRT(A,B,C,D)==true)ans=min(ans,temp);}}}}if(ans==INF) ans=-1;printf("%d\n",ans);}return 0;
}