[校内训练20_01_20]ABC

1.问有多少个大小为N的无标号无根树，直径恰好为L。$N,L leq 200$

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2.问一个竞赛图中有多少个长度为3、4、5的环。$N leq 2000$

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3.给出一些直线和单个点A，问这些直线的交点与A最近的M个距离之和为多少。$N leq 50000,M leq 10^7$。保证不存在两个交点与点A的距离相同。

二分圆的半径，算交点个数，最后统计答案用并查集。

#include<bits/stdc++.h>
using namespace std;
typedef long double ld;
const int maxn=1E5+5;
const ld eps=1E-9;
const ld pi=acos(-1);
const ld inf=1E12;
int n,m;
ld X,Y;
struct pt
{
    ld x,y;
    pt(ld a=0,ld b=0):x(a),y(b){}
    pt operator+(const pt&A){return pt(x+A.x,y+A.y);}
    pt operator-(const pt&A){return pt(x-A.x,y-A.y);}
    pt operator*(const ld&d){return pt(x*d,y*d);}
    pt operator/(const ld&d){return pt(x/d,y/d);}
    ld operator*(const pt&A){return x*A.y-y*A.x;}
    inline void out()
    {
        cout<<"("<<x<<","<<y<<")";
    }
}O;
struct line
{
    pt A,B;
    line(pt a=pt(),pt b=pt()):A(a),B(b){}
}a[maxn];
struct BIT
{
    int tot;
    int t[maxn];
    inline void clear(){memset(t,0,sizeof(t));}
    inline int lowbit(int x){return x&(-x);}
    inline int ask(int x){int sum=0;while(x){sum+=t[x];x-=lowbit(x);}return sum;}
    inline void add(int x,int y){while(x<=tot){t[x]+=y;x+=lowbit(x);}}
}B;
int size;
struct info
{
    int type,num;
    ld x;
    info(int a=0,int b=0,ld c=0):type(a),num(b),x(c){}
    bool operator<(const info&A)const
    {
        return x<A.x;
    }
}tmp[maxn];
inline pt intersection(line a,line b)
{
    pt A=b.B-b.A,B=a.B-a.A,C=b.A-a.A;
    if(abs(A*B)<=eps)
        return pt(inf,inf);
    ld d=-(B*C)/(B*A);
    return b.A+A*d;
}
inline pt T(pt A)
{
    swap(A.x,A.y);
    A.y=-A.y;
    return A;
}
inline pt perpendicular(pt A,line x)
{
    pt B=T(x.B-x.A)+A;
    return intersection(line(A,B),x);
}
inline ld s(ld x)
{
    return x*x;
}
int rk[2][maxn];
inline bool check(ld r)
{
    size=0;
    for(int i=1;i<=n;++i)
    {
        pt A=perpendicular(O,a[i]);
        ld x=s(A.x-O.x)+s(A.y-O.y);
        if(x>=r*r+eps)
            continue;
        pt d=T((O-A)*sqrt(r*r/x-1));
        pt P1=A+d,P2=A-d;
        ld x1=atan2(P1.y-O.y,P1.x-O.x);
        ld x2=atan2(P2.y-O.y,P2.x-O.x);
        if(!(x1<0)&&!(x1>=0))
        {
            x1=atan2(a[i].A.y-a[i].B.y,a[i].A.x-a[i].B.x);
            if(x1<0)
                x2=x1+pi;
            else
                x2=x1-pi;
        }
        if(x1>x2)
            swap(x1,x2);
        tmp[++size]=info(0,i,x1);
        tmp[++size]=info(1,i,x2);
    }
    sort(tmp+1,tmp+size+1);
    for(int i=1;i<=size;++i)
        rk[tmp[i].type][tmp[i].num]=i;
    B.clear();
    B.tot=size;
    long long sum=0;
    for(int i=1;i<=size;++i)
    {
        if(tmp[i].type==0)
            B.add(i,1);
        else
        {
            sum+=B.ask(i-1)-B.ask(rk[0][tmp[i].num]);
            B.add(rk[0][tmp[i].num],-1);
        }
    }
    return sum>=m;
}
int fa[maxn];
int find(int x)
{
    return x==fa[x]?x:fa[x]=find(fa[x]);
}
bool ok[maxn];
inline ld dis(pt A,pt B)
{
    return sqrt(s(A.x-B.x)+s(A.y-B.y));
}
inline ld get(ld r)
{
    size=0;
    for(int i=1;i<=n;++i)
    {
        pt A=perpendicular(O,a[i]);
        ld x=s(A.x-O.x)+s(A.y-O.y);
        if(x>=r*r+eps)
            continue;
        ok[i]=1;
        pt d=T((O-A)*sqrt(r*r/x-1));
        pt P1=A+d,P2=A-d;
        ld x1=atan2(P1.y-O.y,P1.x-O.x);
        ld x2=atan2(P2.y-O.y,P2.x-O.x);
        if(abs(O.x-A.x)<=eps&&abs(O.y-A.y)<=eps)
        {
            x1=atan2(a[i].A.y-a[i].B.y,a[i].A.x-a[i].B.x);
            if(x1<0)
                x2=x1+pi;
            else
                x2=x1-pi;
        }
        if(x1>x2)
            swap(x1,x2);
        tmp[++size]=info(0,i,x1);
        tmp[++size]=info(1,i,x2);
    }
    sort(tmp+1,tmp+size+1);
    for(int i=1;i<=n;++i)
    for(int i=1;i<=size;++i)
        fa[i]=rk[tmp[i].type][tmp[i].num]=i;
    fa[size+1]=size+1;
    ld sum=0;
    for(int i=1;i<=size;++i)
        if(tmp[i].type)
        {
            int l=rk[0][tmp[i].num];
            fa[l]=l+1,fa[i]=i+1;
            while((l=find(l))<i)
            {
                pt A=intersection(a[tmp[l].num],a[tmp[i].num]);
                sum+=dis(intersection(a[tmp[l].num],a[tmp[i].num]),O);
                ++l;
            }
        }
    return sum;
}
inline int R()
{
    return rand()%10000-5000;
}
int main()
{
//    freopen("stigmata.in","r",stdin);
//    freopen("stigmata.out","w",stdout);
    ios::sync_with_stdio(false);
    cin>>n>>X>>Y>>m;
    X/=1000,Y/=1000;
    for(int i=1;i<=n;++i)
    {
        ld x,y;
        cin>>x>>y;
        x/=1000,y/=1000;
        a[i]=line(pt(0,y),pt(1000,1000*x+y));
    }
    O=pt(X,Y);
    ld L=0,R=10000,mid;
    while(abs(L-R)>0.00000001)
    {
        mid=(L+R)/2;
        if(check(mid))
            R=mid;
        else
            L=mid;
    }
    cout<<fixed<<setprecision(7)<<get(R)<<endl;
    return 0;
}