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-9;
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[200005];
int tmp[200005];
int read(){
    int t=0,f=1;char ch=getchar();
    while (ch<'0'||ch>'9'){if (ch=='-')f=-1;ch=getchar();}
    while ('0'<=ch&&ch<='9'){t=t*10+ch-'0';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 0;
    else return 1;
}
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 1;
    return 0;
}
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 0;
    return 1;
}
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=1;
    for (int i=1;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+1,l+1+n,cmp);
    int id=1;
    l[1].id=1;
    for (int i=2;i<=n;i++)
     if (l[i].p!=l[i-1].p)
      l[i].id=++id;
     else
      l[i].id=id; 
    for (int i=1;i<=n;i++)
     for (int j=i+1;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=1;
        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=1;k<=n;k++) tmp[k]=0;
        tmp[l[i].id]=1;tmp[l[j].id]=1;
        for (int k=1;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]=1;
        int cnt=0;
        for (int k=1;k<=n;k++) if (tmp[k]) cnt++;  
        ans=std::max(ans,cnt);  
        cnt=2;
        for (int k=1;k<=n;k++)
         if (k!=i&&k!=j)
         if ((inter(l[k],l[i])==p)) cnt++;
        ans=std::max(ans,cnt);
     }
     printf("%d
",ans);
}
int main(){
    freopen("spider.txt","r",stdin);
    n=read();
    for (int i=1;i<=n;i++){
        int kx=read(),bx=read(),ky=read(),by=read();
        Point p1,p2;
        p1.x=-1;p1.y=-kx+bx;p1.z=-ky+by;
        p2.x=1;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=1;
ll kx[sz],ky[sz],bx[sz],by[sz];
int read(){
    int t=0,f=1;char ch=getchar();
    while (ch<'0'||ch>'9'){if (ch=='-')f=-1;ch=getchar();}
    while ('0'<=ch&&ch<='9'){t=t*10+ch-'0';ch=getchar();}
    return t*f;
}
ll gcd(ll a,ll b){
    if (b==0) 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 0;
      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 1;
  } else return 0;
}
void plane(ll i,ll j,ll &A,ll &B,ll &C,ll &D){
    if (bx[i]-bx[j]==0&&by[i]-by[j]==0){
        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 * 3325443 + b * 3249082 + c * 2308478 + d * 2390850) % 3214567;
    if (s < 0) s = -s;
    return s;
    }
    ll node[3214567], 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 1;
        }
        return 0;
    }
}Point,Plane,Slope;
int main(){
    freopen("spider.txt","r",stdin);
    ll A,B,C,D;
    n=read();
    for (int i=1;i<=n;i++)
     kx[i]=read(),bx[i]=read(),ky[i]=read(),by[i]=read();
    for (int i=1;i<=n;i++)
     for (int j=i+1;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=1;i<=Point.e;i++){
        for (ll a=1;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=1;i<=Plane.e;i++){
        for (ll a=1;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,0)){
                        Plane.pass[i]++;
                        Slope.ins(kx[a],ky[a],i,0);
                    }
        }
    }    
    for (int i=1;i<=Point.e;i++)
     if (Point.pass[i]>ans) ans=Point.pass[i];
    for (int i=1;i<=Plane.e;i++)
     if (Plane.pass[i]>ans) ans=Plane.pass[i]; 
    printf("%d
",ans);  
}