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);
}

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