要点

• 用A、B、C一般式确定每条直线
• 将合法的圆心中点存到每条直线所属的vector中
• 枚举所有线段，二分后\(O(1)\)得到其中存在多少答案，累加

#include <cstdio>
#include <cmath>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <vector>
#include <unordered_map>
using namespace std;

typedef long long ll;
const int maxn = 3e5 + 5, maxm = 1550;
const ll inf = 1e18;

struct Point {
    ll x, y;
    Point() {}
    Point(ll a, ll b):x(a), y(b) {}
};
struct Circle {
    Point p;
    ll r;
    Circle() {}
    Circle(Point a, ll b):p(a), r(b) {}
};

int n, m, hashcnt;
Point a[maxn], b[maxn];//segments
Circle c[maxm];//circles
unordered_map<ll, int> mp;//<{A, B, C}, id>
vector<ll> v[maxn];//v[id]
ll A, B, C;//a few times used
ll ans;
int ID[maxn];

ll sqr(ll x) {
    return x * x;
}

ll dis(int i, int j) {//distance ^ 2
    return sqr(c[i].p.x - c[j].p.x) + sqr(c[i].p.y - c[j].p.y);
}

ll gcd(ll a, ll b) {//exist zero: return
    if (!a || !b)   return a + b;
    return gcd(b, a % b);
}

ll Cross(Point A, Point B) {
    return A.x * B.y - A.y * B.x;
}

void Deal(ll &A, ll &B, ll &C) {//unique the line
    ll q = gcd(gcd(abs(A), abs(B)), abs(C));
    A /= q, B /= q, C /= q;
    if (A == 0 && B < 0)    B = -B, C = -C; 
    else if (A < 0) A = -A, B = -B, C = -C;
}

ll hashi(ll A, ll B, ll C) {//map is TLE, so I hash it
    return A * (ll)(1e12) + B * (ll)(1e6) + C;
}

void Hash(int i, ll A, ll B, ll C) {
    ll d = hashi(A, B, C);
    if (!mp.count(d)) {
        mp[d] = ++hashcnt;
    }
    int id = mp[d];
    ID[i] = id;
}

void calc(Point a, Point b) {// get A, B, C
    A = b.y - a.y, B = a.x - b.x;
    C = Cross(b, a);//Ax + By + C = 0   
    Deal(A, B, C);
}

int main() {
    scanf("%d %d", &n, &m);
    for (int i = 1; i <= n; i++) {
        scanf("%lld %lld %lld %lld", &a[i].x, &a[i].y, &b[i].x, &b[i].y);
        a[i].x *= 2, a[i].y *= 2, b[i].x *= 2, b[i].y *= 2;//for line 99
        if (a[i].x > b[i].x)    swap(a[i], b[i]);//for line 109 & 110
        calc(a[i], b[i]);
        Hash(i, A, B, C);
    }
    for (int i = 1; i <= m; i++) {
        scanf("%lld %lld %lld", &c[i].p.x, &c[i].p.y, &c[i].r);
        c[i].p.x *= 2, c[i].p.y *= 2, c[i].r *= 2;
        for (int j = 1; j < i; j++)
            if (c[i].r == c[j].r && dis(i, j) > 4LL * sqr(c[i].r)) {//if eyes
                calc(c[i].p, c[j].p);
                ll A1 = B * 2, B1 = -A * 2;
                ll C1 = A * (c[i].p.y + c[j].p.y) - B * (c[i].p.x + c[j].p.x);
                Deal(A1, B1, C1);
                ll d = hashi(A1, B1, C1);
                if (!mp.count(d))   continue;
                ll x = (c[i].p.x + c[j].p.x) / 2;//line 99: should avoid double
                v[mp[d]].emplace_back(x);
            }
    }
    for (int i = 1; i <= hashcnt; i++) {
        v[i].emplace_back(inf);
        sort(v[i].begin(), v[i].end());
    }
    for (int i = 1; i <= n; i++) {
        int id = ID[i];
        int l = lower_bound(v[id].begin(), v[id].end(), a[i].x) - v[id].begin();//line 109
        int r = upper_bound(v[id].begin(), v[id].end(), b[i].x) - v[id].begin();//line 110
        ans += r - l;
    }
    return !printf("%lld\n", ans);
}