根据初中数学知识，圆心在 \(AB\) 线段的中垂线上。
又因为给定圆与 \(AB\) 线段所在直线不交，所以圆心在中垂线的一端极远处完全包含这个给定圆，在另一端极远处与这个给定圆相离。而具体在哪一端只与圆心在 \(AB\) 的左侧还是右侧有关。
因此可以二分找到与给定圆外切和内切时圆心的坐标。那么一个给定圆只可能挡住一段连续的坐标。扫描线即可。
code
#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mkp make_pair
#define mems(a, x) memset((a), (x), sizeof(a))
using namespace std;
typedef long long ll;
typedef double db;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<ll, ll> pii;
const int maxn = 100100;
ll A, B, C, D, n, a[maxn], b[maxn], c[maxn];
ldb mx, my;
inline ldb calc(ldb x, ldb y) {
return sqrtl(x * x + y * y);
}
void solve() {
scanf("%lld%lld%lld%lld%lld", &A, &B, &C, &D, &n);
mx = (ldb)(A + C) / 2;
my = (ldb)(B + D) / 2;
vector< pair<ldb, int> > vc;
for (int i = 1; i <= n; ++i) {
scanf("%lld%lld%lld", &a[i], &b[i], &c[i]);
}
if (B == D) {
swap(mx, my);
swap(A, B);
swap(C, D);
for (int i = 1; i <= n; ++i) {
swap(a[i], b[i]);
}
}
ldb kk = -(ldb)(A - C) / (B - D);
ldb bb = my - mx * kk;
ldb k1 = (ldb)(B - D) / (A - C);
ldb b1 = my - mx * k1;
for (int i = 1; i <= n; ++i) {
if ((A != C && (b[i] - b1) / k1 > a[i]) || (A == C && A > a[i])) {
// if (0) {
ldb l = -2e12, r = 2e12;
while (r - l > 1e-3) {
ldb mid = (l + r) / 2;
ldb x = mid, y = kk * mid + bb;
ldb dis = calc(x - a[i], y - b[i]), R = calc(A - x, B - y);
// printf("%.5Lf %.5Lf %.5Lf %lld %.5Lf\n", x, y, dis, c[i], R);
if (dis + c[i] <= R) {
l = mid;
} else {
r = mid;
}
}
// printf("%.5Lf ", l);
vc.pb(l, 1);
r = 2e12;
while (r - l > 1e-3) {
ldb mid = (l + r) / 2;
ldb x = mid, y = kk * mid + bb;
ldb dis = calc(x - a[i], y - b[i]), R = calc(A - x, B - y);
// printf("%.5Lf %.5Lf %.5Lf %lld %.5Lf\n", x, y, dis, c[i], R);
if (dis <= R + c[i]) {
l = mid;
} else {
r = mid;
}
}
vc.pb(l, -1);
// printf("%.5Lf\n", l);
} else {
// printf("i: %d\n", i);
ldb l = -2e12, r = 2e12;
while (r - l > 1e-3) {
ldb mid = (l + r) / 2;
ldb x = mid, y = kk * mid + bb;
ldb dis = calc(x - a[i], y - b[i]), R = calc(A - x, B - y);
// printf("%.5Lf %.5Lf %.5Lf %lld %.5Lf\n", x, y, dis, c[i], R);
if (dis + c[i] >= R) {
l = mid;
} else {
r = mid;
}
}
// printf("%.5Lf ", l);
vc.pb(l, -1);
l = -2e12;
while (r - l > 1e-3) {
ldb mid = (l + r) / 2;
ldb x = mid, y = kk * mid + bb;
ldb dis = calc(x - a[i], y - b[i]), R = calc(A - x, B - y);
// printf("%.5Lf %.5Lf %.5Lf %lld %.5Lf\n", x, y, dis, c[i], R);
if (dis >= R + c[i]) {
l = mid;
} else {
r = mid;
}
}
vc.pb(l, 1);
// printf("%.5Lf\n", l);
}
}
vc.pb(mx, 0);
sort(vc.begin(), vc.end());
int x = 0;
ldb ans = 1e18;
for (auto p : vc) {
if (!x) {
// printf("x, y: %.5Lf %.5Lf\n", p.fst, kk * p.fst + bb);
ans = min(ans, calc(A - p.fst, B - (kk * p.fst + bb)));
}
x += p.scd;
if (!x) {
// printf("x, y: %.5Lf %.5Lf\n", p.fst, kk * p.fst + bb);
ans = min(ans, calc(A - p.fst, B - (kk * p.fst + bb)));
}
}
printf("%.5Lf\n", ans);
}
int main() {
int T = 1;
// scanf("%d", &T);
while (T--) {
solve();
}
return 0;
}