FJ省队集训DAY2 T1

思路：转换成n条三维空间的直线，求最大的集合使得两两有交点。

有两种情况:第一种是以某2条直线为平面，这时候只要统计这个平面上有几条斜率不同的直线就可以了

还有一种是全部交于同一点，这个也只要判断就可以了。

然后我并不能改出来，wa了好多个点

WA的程序:

 #include<cstdio>
 #include<iostream>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 #define dou long double
 const dou eps=1e-;
 int n;
 struct Point{
     dou x,y,z;
     Point(){}
     Point(dou x0,dou y0,dou z0):x(x0),y(y0),z(z0){}
 };
 struct Line{
     Point s,e,p;
     int id;
     Line(){}
     Line(Point s0,Point e0):s(s0),e(e0){}
 }l[];
 int tmp[];
 int read(){
     int t=,f=;char ch=getchar();
     while (ch<''||ch>''){if (ch=='-')f=-;ch=getchar();}
     while (''<=ch&&ch<=''){t=t*+ch-'';ch=getchar();}
     return t*f;
 }
 bool cmp(Line l1,Line l2){
     Point p1=l1.p,p2=l2.p;
     if (fabs(p1.x-p2.x)<eps&&fabs(p1.y-p2.y)<eps) return p1.z<p2.z;
     else
     if (fabs(p1.x-p2.x)<eps) return p1.y<p2.y;
     return p1.x<p2.x;
 }
 bool operator ==(Point p1,Point p2){
     return fabs(p1.x-p2.x)<=eps&&fabs(p1.y-p2.y)<=eps&&fabs(p1.z-p2.z)<=eps;
 }
 bool operator !=(Point p1,Point p2){
     if (p1==p2) return ;
     else return ;
 }
 double operator /(Point p1,Point p2){
     return p1.x*p2.x+p1.y*p2.y+p1.z*p2.z;
 }
 Point operator -(Point p1,Point p2){
     return Point(p1.x-p2.x,p1.y-p2.y,p1.z-p2.z);
 }
 Point operator *(Point p1,Point p2){
     return Point(p1.y*p2.z-p1.z*p2.y,p1.z*p2.x-p1.x*p2.z,p1.x*p2.y-p1.y*p2.x);
 }
 Point operator /(Point p1,dou x){
     return Point(p1.x/x,p1.y/x,p1.z/x);
 }
 dou sqr(dou x){
     return x*x;
 }
 dou dist(Point p){
     return sqrt(sqr(p.x)+sqr(p.y)+sqr(p.z));
 }
 dou dist(Point p1,Point p2){
     return dist(p1-p2);
 }
 bool zero(dou x){
     if (fabs(x)<eps) return ;
     return ;
 }
 bool zero(Point p){
     return zero(p.x)&&zero(p.y)&&zero(p.z);
 }
 bool LineIntersect(Line p1, Line p2){
     dou x1=p1.s.x,x2=p1.e.x,x3=p2.s.x,x4=p2.e.x;
     dou y1=p1.s.y,y2=p1.e.y,y3=p2.s.y,y4=p2.e.y;
     dou z1=p1.s.z,z2=p1.e.z,z3=p2.s.z,z4=p2.e.z;
     dou x12=(x1-x2),x13=(x1-x3),x34=(x3-x4);
     dou y12=(y1-y2),y13=(y1-y3),y34=(y3-y4);
     dou z12=(z1-z2);
     dou t=(y34*x12-x34*y12);
     if (fabs(t)<eps) return ;
     return ;
 }
 Point inter(Line p1,Line p2){
     dou x1=p1.s.x,x2=p1.e.x,x3=p2.s.x,x4=p2.e.x;
     dou y1=p1.s.y,y2=p1.e.y,y3=p2.s.y,y4=p2.e.y;
     dou z1=p1.s.z,z2=p1.e.z,z3=p2.s.z,z4=p2.e.z;
     dou x12=(x1-x2),x13=(x1-x3),x34=(x3-x4);
     dou y12=(y1-y2),y13=(y1-y3),y34=(y3-y4);
     dou z12=(z1-z2);
     dou t=(y13*x34-y34*x13)/(y34*x12-x34*y12);
     dou x=x1+x12*((y13*x34-y34*x13)/(y34*x12-x34*y12));
     dou y=y1+y12*t,z=z1+z12*t;
     return Point(x,y,z);
 }
 void solve(){
     int ans=;
     for (int i=;i<=n;i++){
         Point p=l[i].e-l[i].s;
         dou len=dist(p);
         p=p/len;
         l[i].p=p;
     }
     std::sort(l+,l++n,cmp);
     int id=;
     l[].id=;
     for (int i=;i<=n;i++)
      if (l[i].p!=l[i-].p)
       l[i].id=++id;
      else
       l[i].id=id;
     for (int i=;i<=n;i++)
      for (int j=i+;j<=n;j++)
       if (LineIntersect(l[i],l[j])){
         Point p=inter(l[i],l[j]);
         Point e1=l[i].e-l[i].s,e2=l[j].e-l[j].s;
         dou len1=dist(e1),len2=dist(e2);
         e1=e1/len1;e2=e2/len2;
         dou x1=e1.x,x2=e2.x,y1=e1.y,y2=e2.y,z1=e1.z,z2=e2.z;
         dou x=;
         dou y=(x2*z1-x1*z2)/(y1*z2-z1*y2);
         dou z=(-x1*x-y1*y)/z1;
         Point e=Point(x,y,z);
         for (int k=;k<=n;k++) tmp[k]=;
         tmp[l[i].id]=;tmp[l[j].id]=;
         for (int k=;k<=n;k++)
          if (k!=i&&k!=j)
           if (fabs((l[k].e-l[k].s)/e)<eps&&LineIntersect(l[i],l[k])) tmp[l[k].id]=;
         int cnt=;
         for (int k=;k<=n;k++) if (tmp[k]) cnt++;
         ans=std::max(ans,cnt);
         cnt=;
         for (int k=;k<=n;k++)
          if (k!=i&&k!=j)
          if ((inter(l[k],l[i])==p)) cnt++;
         ans=std::max(ans,cnt);
      }
      printf("%d\n",ans);
 }
 int main(){
     freopen("spider.txt","r",stdin);
     n=read();
     for (int i=;i<=n;i++){
         int kx=read(),bx=read(),ky=read(),by=read();
         Point p1,p2;
         p1.x=-;p1.y=-kx+bx;p1.z=-ky+by;
         p2.x=;p2.y=kx+bx;p2.z=ky+by;
         l[i]=Line(p1,p2);
     }
     solve();
 }

只好改成std的写法了。

 #include<cstdio>
 #include<iostream>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 #define sz 1020000
 #define ll long long int
 using namespace std;
 int n,ans=;
 ll kx[sz],ky[sz],bx[sz],by[sz];
 int read(){
     int t=,f=;char ch=getchar();
     while (ch<''||ch>''){if (ch=='-')f=-;ch=getchar();}
     while (''<=ch&&ch<=''){t=t*+ch-'';ch=getchar();}
     return t*f;
 }
 ll gcd(ll a,ll b){
     if (b==) return a;
     else return gcd(b,a%b);
 }
 bool ok(ll i,ll j,ll &X,ll &Y,ll &T,ll &T2){
     if (kx[i] == kx[j]){
       if (ky[i] == ky[j]) return ;
       T = by[i] - by[j]; T2 = ky[j] - ky[i];
     } else{
       T = bx[i] - bx[j]; T2 = kx[j] - kx[i];
     }
     if (kx[i] * T + bx[i] * T2 == kx[j] * T + bx[j] * T2 &&
     ky[i] * T + by[i] * T2 == ky[j] * T + by[j] * T2){
     X = kx[i] * T + bx[i] * T2;
     Y = ky[i] * T + by[i] * T2;
     ll g = gcd(gcd(X, Y), gcd(T, T2));
     X /= g; Y /= g; T /= g; T2 /= g;

     return ;
   } else return ;
 }
 void plane(ll i,ll j,ll &A,ll &B,ll &C,ll &D){
     if (bx[i]-bx[j]==&&by[i]-by[j]==){
         A=ky[i]-ky[j],B=kx[i]-kx[j];
     }else{
         A=by[i]-by[j],B=bx[i]-bx[j];
     }
     C=A*kx[i]+B*ky[i];
     D=A*bx[i]+B*by[i];
     ll g=gcd(gcd(A,B),gcd(C,D));
     A/=g;B/=g;C/=g;D/=g;
 }
 class Hash{
 public:
     ll F(ll a, ll b, ll c, ll d){
     ll s = (a *  + b *  + c *  + d * ) % ;
     if (s < ) s = -s;
     return s;
     }
     ll node[], next[sz], A[sz], B[sz], C[sz], D[sz], pass[sz];
     ll e;
     void ins(ll a,ll b,ll c,ll d){
         ll s=F(a,b,c,d);
         e++;
         next[e]=node[s];node[s]=e;
         A[e]=a;B[e]=b;C[e]=c;D[e]=d;
     }
     bool find(ll a,ll b,ll c,ll d){
         ll s=F(a,b,c,d),j;
         for (j=node[s];j;j=next[j]){
             if (A[j]==a&&B[j]==b&&C[j]==c&&D[j]==d)
             return ;
         }
         return ;
     }
 }Point,Plane,Slope;
 int main(){
     freopen("spider.txt","r",stdin);
     ll A,B,C,D;
     n=read();
     for (int i=;i<=n;i++)
      kx[i]=read(),bx[i]=read(),ky[i]=read(),by[i]=read();
     for (int i=;i<=n;i++)
      for (int j=i+;j<=n;j++)
       if (ok((ll)i,(ll)j,A,B,C,D)){
             if (!Point.find(A,B,C,D)) Point.ins(A,B,C,D);
             plane(i,j,A,B,C,D);
             if (!Plane.find(A,B,C,D)) Plane.ins(A,B,C,D);
       }
     for (ll i=;i<=Point.e;i++){
         for (ll a=;a<=n;a++){
             if (kx[a]*Point.C[i]+bx[a]*Point.D[i]==Point.A[i])
              if (ky[a]*Point.C[i]+by[a]*Point.D[i]==Point.B[i])
               Point.pass[i]++;
         }
     }
     for (ll i=;i<=Plane.e;i++){
         for (ll a=;a<=n;a++){
             if (kx[a] * Plane.A[i] + ky[a] * Plane.B[i] == Plane.C[i])
                 if (bx[a] * Plane.A[i] + by[a] * Plane.B[i] == Plane.D[i])
                     if (!Slope.find(kx[a],ky[a],i,)){
                         Plane.pass[i]++;
                         Slope.ins(kx[a],ky[a],i,);
                     }
         }
     }
     for (int i=;i<=Point.e;i++)
      if (Point.pass[i]>ans) ans=Point.pass[i];
     for (int i=;i<=Plane.e;i++)
      if (Plane.pass[i]>ans) ans=Plane.pass[i];
     printf("%d\n",ans);
 }