HNOI2019 鱼 fish

本来想写个改题记录的然后想了想改不完所以就分开写了= =

https://www.luogu.org/problemnew/show/P5286

显然枚举A，D，然后鱼头和鱼身分开来考虑。

鱼身：先枚举B,C，那么BC的中点一定在线段AD（不包含端点）上，对于每一条直线维护一个vector存所有的点，将这个BC的中点插入进线段BC的垂直平分线的vector，然后对于一组AD，鱼身的方案数是vector上AD中间的点数，可以用upper_bound求。

鱼尾：枚举D，对所有其他点极角排序，用双指针扫一遍所有点即可。具体难以描述请直接看代码

这题细节巨多，考场上一看就会写，写了3h还没拍上以为暴力写挂走了，结果水到40分，如果没被卡常有60，然后今天又调了一个上午才过的= =说不定还有bug只是数据太水= =

// luogu-judger-enable-o2
#include<bits/stdc++.h>
#define il inline
#define rg register
#define vd void
#define ll long long
il int gi(){
    int x=0,f=0;char ch=getchar();
    while(!isdigit(ch))f^=ch=='-',ch=getchar();
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    return f?-x:x;
}
int n;
template<class T>il T gcd(T a,T b){return b?gcd(b,a%b):a;}
struct number{
    ll x,y;
    number(){}
    number(const ll&a,const ll&b){
        ll g=gcd(llabs(a),llabs(b));
        x=a/g,y=b/g;
        if(y<0)x=-x,y=-y;
    }
};
il bool operator<(const number&a,const number&b){
    if(a.x^b.x)return a.x<b.x;
    return a.y<b.y;
}
il bool operator!=(const number&a,const number&b){return a.x!=b.x||a.y!=b.y;}
struct line{number k,b;};
il bool operator<(const line&a,const line&b){
    if(a.k!=b.k)return a.k<b.k;
    return a.b<b.b;
}
std::map<line,int>M;int cnt;
int x[1010],y[1010];
struct yyb{int i;double at2;}s[2010];
il bool operator<(const yyb&a,const yyb&b){return a.at2<b.at2;}
std::vector<int>vec[1000010];
il ll getdist(int a,int b){
    return 1ll*(x[a]-x[b])*(x[a]-x[b])+1ll*(y[a]-y[b])*(y[a]-y[b]);
}
int tail[1010][1010];
std::map<ll,int>mmp;
int res,nowi;
const double eps=1e-10,pi=acos(-1);
il vd insert(int x){res+=mmp[getdist(nowi,x)]++;}
il vd delet(int x){res-=--mmp[getdist(nowi,x)];}
int main(){
    //freopen("fish.in","r",stdin);
    //freopen("fish.out","w",stdout);
    n=gi();
    for(int i=1;i<=n;++i)x[i]=gi(),y[i]=gi();
    for(int i=1;i<=n;++i)
        for(int j=i+1;j<=n;++j)
            if(y[i]==y[j]){
                if(x[i]+x[j]&1)continue;
                number k=(number){1,0},b=(number){(x[i]+x[j])/2,1};
                if(M.find((line){k,b})==M.end())M[(line){k,b}]=++cnt;
                vec[M[(line){k,b}]].push_back(y[i]*2);
            }else{
                number k=(number){x[i]-x[j],y[j]-y[i]},b=(number){-1ll*(x[i]-x[j])*(x[i]+x[j])+1ll*(y[j]-y[i])*(y[j]+y[i]),2ll*(y[j]-y[i])};
                if(M.find((line){k,b})==M.end())M[(line){k,b}]=++cnt;
                vec[M[(line){k,b}]].push_back(x[i]==x[j]?x[i]*2:y[i]+y[j]);
            }
    for(int i=1;i<=n;++i){
        int m=0;
        for(int j=1;j<=n;++j)if(j!=i)s[++m]=(yyb){j,atan2(y[j]-y[i],x[j]-x[i])};
        std::sort(s+1,s+m+1);
        for(int i=1;i<=m;++i)s[i+m]=s[i],s[i+m].at2+=pi*2;
        mmp.clear();
        res=0;nowi=i;
        for(int j=1,p=0,q=0;j<=m;++j){
            while(s[p+1].at2+eps<s[j].at2+1.5*pi)insert(s[++p].i);
            while(s[q+1].at2<s[j].at2+0.5*pi+eps)delet(s[++q].i);
            tail[s[j].i][i]=res;
        }
    }
    for(int i=1;i<=cnt;++i)std::sort(vec[i].begin(),vec[i].end());
    ll ans=0;
    for(int i=1;i<=n;++i)
        for(int j=i+1;j<=n;++j){
            int xa=y[i],xb=y[j];
            if(xa==xb)xa=x[i],xb=x[j];
            if(xa>xb)std::swap(xa,xb);
            number k,b;
            if(x[i]==x[j])k=(number){1,0},b=(number){x[i],1};
            else k=(number){y[i]-y[j],x[i]-x[j]},b=(number){-1ll*x[i]*(y[i]-y[j])+1ll*y[i]*(x[i]-x[j]),x[i]-x[j]};
            if(M.find((line){k,b})==M.end())continue;
            int veci=M[(line){k,b}];
            ans+=(std::upper_bound(vec[veci].begin(),vec[veci].end(),xb*2-1)-std::upper_bound(vec[veci].begin(),vec[veci].end(),xa*2))*(tail[i][j]+tail[j][i]);
        }
    printf("%lld\n",ans*4);
    return 0;
}