Codeforces Round #203 - D. Looking for Owls

D. Looking for Owls

Emperor Palpatine loves owls very much. The emperor has some blueprints with the new Death Star, the blueprints contain n distinct segments and m distinct circles. We will consider the segments indexed from 1 to n in some way and the circles — indexed from 1 to m in some way.

Palpatine defines an owl as a set of a pair of distinct circles (i, j) (i < j) and one segment k, such that:

circles i and j are symmetrical relatively to the straight line containing segment k;
circles i and j don't have any common points;
circles i and j have the same radius;
segment k intersects the segment that connects the centers of circles i and j.

Help Palpatine, count the number of distinct owls on the picture.

Input

The first line contains two integers — n and m (1 ≤ n ≤ 3·10⁵, 2 ≤ m ≤ 1500).

The next n lines contain four integers each, x₁, y₁, x₂, y₂ — the coordinates of the two endpoints of the segment. It's guaranteed that each segment has positive length.

The next m lines contain three integers each, x_i, y_i, r_i — the coordinates of the center and the radius of the i-th circle. All coordinates are integers of at most 10⁴ in their absolute value. The radius is a positive integer of at most 10⁴.

It is guaranteed that all segments and all circles are dictinct.

Output

Print a single number — the answer to the problem.

Please, do not use the %lld specifier to output 64-bit integers is С++. It is preferred to use the cout stream or the %I64d specifier.

Sample test(s)

Input

Output

Input

Output

Input

Output

Note

Here's an owl from the first sample. The owl is sitting and waiting for you to count it.

Codeforces Round #203 - D. Looking for Owls

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题意：有n条线段和m个圆（1 ≤ n ≤ 3·10⁵, 2 ≤ m ≤ 1500），如果存在两个圆和这两个圆的对称线段（并且线段要和圆心的连线相交），那么一起称为一个owl，求一共有多少个owl。

思路：最暴力的做法就是枚举每一对圆，求它们的对称线，然后枚举每一条线段，判断线段是否在对称线上，复杂度o(m²n)，但是数据规模太大无法承受，这题不是一道纯粹的几何题。

后来我想，如果在暴力的基础上改进一下，先预处理出每对圆的对称线，在枚举每一条线段的时候，如果能在o(1)或o(logn)内判断出这条对称线是否存在，应该能过。

于是我想到了将一条直线hash成一个long long值，这种hash还是第一次写，我将这条直线对应的向量通过求gcd将x，y都缩到最小，并且保证x非负，然后在这条直线上取一点使得x非负并且尽量小，将这向量和这点的坐标值随便乘一些很大的数再加起来就算hash了，这样就可以保证同一条直线不管已知哪两点，算出的hash值都唯一。

现在问题来了，怎么判断线段和圆心连线相交呢？如果再找出这两个圆心的话，那跟暴力无异。于是我又想到了将一个对称线的hash值和对称线和圆心的交点的x坐标弄成一个pair插入到一个vector中，全部完了后对vector排序，hash值小的在前，相等的话x坐标小的在前面，然后枚举每一条线段，将两端点和对应的向量也都hash一下然后弄成pair，对应用lower_bound和upper_bound二分查找，在这个区间内的pair肯定都是符合的，ans+=区间长度就可以了。

于是时间复杂度降到了o(m²+nlogm²)。

 #include <iostream>

 #include <stdio.h>

 #include <map>

 #include <cmath>

 #include <vector>

 #include <algorithm>

 using namespace std;

 #define TR(x) (x+10000)*2

 typedef long long ll;

 struct Point

 {

     int x, y;

     Point(int x=, int y=):x(x),y(y) { }

 };

 typedef Point Vector;

 Vector operator - (const Point& A, const Point& B)

 {

     return Vector(A.x-B.x, A.y-B.y);

 }

 struct Circle

 {

     int x,y,r;

 };

 struct Line

 {

     Point begin;

     Point end;

 };

 int gcd(int a,int b)

 {

     return b== ? a : gcd(b, a%b);

 }

 Vector simpleVector(Vector v)

 {

     if(v.x==)

         return Vector(,);

     if(v.y==)

         return Vector(,);

     int g=gcd(abs(v.x), abs(v.y));

     if(v.x>)

         return Vector(v.x/g, v.y/g);

     return Vector(-v.x/g, -v.y/g);

 }

 Point simpleCenter(Point p, Vector v) // v.x>=0 && p.x>=0 && p.y>=0

 {

     if(v.x==)

         return Point(p.x,);

     if(v.y==)

         return Point(,p.y);

     Point r;

     r.x = p.x % v.x;

     int k = (p.x-r.x)/v.x;

     r.y=p.y-k*v.y;

     return r;

 }

 ll hash(Point p, Vector v)

 {

     ll ans= (ll)p.x*80001LL

             +(ll)p.y*80001LL*

             +(ll)v.x*80001LL**

             +(ll)v.y*80001LL***;

     return ans;

 }

 typedef pair<ll, int> Pair;

 vector<Pair> a;

 Circle cs[];

 Line ls[];

 int main()

 {

     // 坐标+10000，再放大2倍，坐标范围[0,40000]，且都是偶数

     freopen("in.txt","r",stdin);

     int nLine, nCircle;

     cin>> nLine>> nCircle;

     for(int i=; i<nLine; i++)

     {

         int x1,y1,x2,y2;

         scanf("%d %d %d %d",&x1,&y1,&x2,&y2);

         ls[i]=Line {Point(TR(x1),TR(y1)),Point(TR(x2),TR(y2))};

     }

     for(int i=; i<nCircle; i++)

     {

         int x,y,r;

         scanf("%d %d %d", &x, &y, &r);

         cs[i]=Circle {TR(x),TR(y),r*};

     }

     for(int i=; i<nCircle; i++)

         for(int j=i+; j<nCircle; j++)

             if(cs[i].r == cs[j].r)

             {

                 Vector c1c2=Vector(cs[j].x-cs[i].x, cs[j].y-cs[i].y);

                 // 判断两个圆是否相交

                 if(c1c2.x*c1c2.x+c1c2.y*c1c2.y <= *cs[i].r*cs[i].r)

                     continue;

                 Vector v=Vector(-cs[j].y+cs[i].y, cs[j].x-cs[i].x); // 垂直向量

                 v=simpleVector(v); // v.x>=0

                 Point center=Point((cs[j].x+cs[i].x)/, (cs[j].y+cs[i].y)/);

                 center = simpleCenter(center, v);

                 a.push_back(Pair(hash(center, v), (cs[j].x+cs[i].x)/));

             }

     sort(a.begin(), a.end());

     int ans=;

     for(int i=; i<nLine; i++)

     {

         Vector v=simpleVector(ls[i].begin-ls[i].end);

         Point p=simpleCenter(ls[i].begin,v);

         ll h = hash(p,v);

         // 在vector内二分查找在区间[L,R]内的的个数

         Pair L = Pair(h, min(ls[i].begin.x, ls[i].end.x));

         Pair R = Pair(h, max(ls[i].begin.x, ls[i].end.x));

         vector<Pair>::iterator i1 = lower_bound(a.begin(), a.end(), L);

         vector<Pair>::iterator i2 = upper_bound(a.begin(), a.end(), R);

         ans+=i2-i1;

     }

     cout<< ans;

     return ;

 }

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Codeforces Round #203 - D. Looking for Owls

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