Codeforces 350D（计算几何）

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);
}