CF886F Symmetric Projections

题意：

标程：

 1 #include<bits/stdc++.h>
 2 #define P pair<int,int>
 3 using namespace std;
 4 const int inf=0x3f3f3f3f;
 5 const double eps=1e-8;
 6 const int N=2005;
 7 struct node{double x,y,k;int px,py;}a[N*N];
 8 double mid_x,mid_y;
 9 int n,x[N],y[N],ans,k,t,p[N],die[N],X[N],Y[N],cnt,lst,pos;
10 bool cmp (const node &A,const node &B) {return A.k<B.k;}
11 bool operator == (const node &A,const node &B) {return fabs(A.k-B.k)<eps;}
12 bool check(int x,double b1,double b2)//double不要开成int！ 
13 {
14     double dx=mid_y-Y[x],dy=-(mid_x-X[x]);
15     return fabs(dy*b1-b2*dx)<eps;
16 }
17 int main()
18 {
19     scanf("%d",&n);
20     for (int i=1;i<=n;i++) 
21       scanf("%d%d",&x[i],&y[i]),mid_x+=x[i],mid_y+=y[i]; 
22     mid_x/=n;mid_y/=n;
23     for (int i=1;i<=n;i++)
24     if (!die[i])
25       for (int j=1;j<=n;j++)
26         if (!die[j]&&fabs(x[i]+x[j]-2*mid_x)<eps&&fabs(y[i]+y[j]-2*mid_y)<eps) {die[i]=1,die[j]=1;break;}
27     for (int i=1;i<=n;i++) if (!die[i]) X[++t]=x[i],Y[t]=y[i];
28     if (t<2) return puts("-1"),0;
29     for (int i=1;i<t;i++)
30       for (int j=i+1;j<=t;j++)
31       {
32            double dx=mid_y-(Y[i]+Y[j])/2.0,dy=-(mid_x-(X[i]+X[j])/2.0);//垂直，斜率负倒数
33            if (dx<0) dx=-dx,dy=-dy; 
34            if (dx==0) dy=fabs(dy);
35            if (dy==0) dx=fabs(dx);//!!!
36          a[++cnt].x=dx,a[cnt].y=dy;a[cnt].px=i;a[cnt].py=j;a[cnt].k=atan2(dy,dx);
37       }
38     sort(a+1,a+cnt+1,cmp);lst=1;
39     for (int i=1;i<=cnt;i++)//check and count 
40       if (i==cnt||!(a[i]==a[i+1]))
41       {
42             if (i-lst+1>=t/2) 
43             {
44                 for (int j=1;j<=t;j++) p[j]=1;
45                 for (;lst<=i;lst++) 
46                   if (p[a[lst].px]&&p[a[lst].py]) p[a[lst].px]=p[a[lst].py]=0;//注意重合的两点不能同时和另一点匹配！一配一 
47                 k=0;
48                 for (int j=1;j<=t;j++) if (p[j]) k++,pos=j;
49                 if (k>(t&1)) continue; 
50                 if (!k||check(pos,a[i].x,a[i].y)) ans++;
51             } 
52             lst=i+1; 
53       }
54     printf("%d\n",ans);
55    return 0;
56 }