[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+;
 const ld eps=1E-;
 const ld pi=acos(-);
 const ld inf=1E12;
 int n,m;
 ld X,Y;
 struct pt
 {
     ld x,y;
     pt(ld a=,ld b=):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,,sizeof(t));}
     inline int lowbit(int x){return x&(-x);}
     inline int ask(int x){int sum=;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=,int b=,ld c=):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[][maxn];
 inline bool check(ld r)
 {
     size=;
     for(int i=;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-));
         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<)&&!(x1>=))
         {
             x1=atan2(a[i].A.y-a[i].B.y,a[i].A.x-a[i].B.x);
             if(x1<)
                 x2=x1+pi;
             else
                 x2=x1-pi;
         }
         if(x1>x2)
             swap(x1,x2);
         tmp[++size]=info(,i,x1);
         tmp[++size]=info(,i,x2);
     }
     sort(tmp+,tmp+size+);
     for(int i=;i<=size;++i)
         rk[tmp[i].type][tmp[i].num]=i;
     B.clear();
     B.tot=size;
     long long sum=;
     for(int i=;i<=size;++i)
     {
         if(tmp[i].type==)
             B.add(i,);
         else
         {
             sum+=B.ask(i-)-B.ask(rk[][tmp[i].num]);
             B.add(rk[][tmp[i].num],-);
         }
     }
     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=;
     for(int i=;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]=;
         pt d=T((O-A)*sqrt(r*r/x-));
         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<)
                 x2=x1+pi;
             else
                 x2=x1-pi;
         }
         if(x1>x2)
             swap(x1,x2);
         tmp[++size]=info(,i,x1);
         tmp[++size]=info(,i,x2);
     }
     sort(tmp+,tmp+size+);
     for(int i=;i<=n;++i)
     for(int i=;i<=size;++i)
         fa[i]=rk[tmp[i].type][tmp[i].num]=i;
     fa[size+]=size+;
     ld sum=;
     for(int i=;i<=size;++i)
         if(tmp[i].type)
         {
             int l=rk[][tmp[i].num];
             fa[l]=l+,fa[i]=i+;
             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()%-;
 }
 int main()
 {
 //    freopen("stigmata.in","r",stdin);
 //    freopen("stigmata.out","w",stdout);
     ios::sync_with_stdio(false);
     cin>>n>>X>>Y>>m;
     X/=,Y/=;
     for(int i=;i<=n;++i)
     {
         ld x,y;
         cin>>x>>y;
         x/=,y/=;
         a[i]=line(pt(,y),pt(,*x+y));
     }
     O=pt(X,Y);
     ld L=,R=,mid;
     while(abs(L-R)>0.00000001)
     {
         mid=(L+R)/;
         if(check(mid))
             R=mid;
         else
             L=mid;
     }
     cout<<fixed<<setprecision()<<get(R)<<endl;
     return ;
 }