10.6 总结

T1

一道计几，还行，第一个就是直接三分支线上的点然后求函数谷值，第二个就是 \(\min\{Dist(x_1, x_3), Dist(x_2, x_3)\}\)。

#include <cmath>
#include <iomanip>
#include <fstream>
#include <ctime>

using namespace std;

const double eps = 1e-8;

ifstream cin("fountain.in");
ofstream cout("fountain.out");

struct Point {
  double x, y;
};

double r;
Point x[3];

double Dis(Point x, Point y) {
  return sqrt(pow(x.x - y.x, 2) + pow(x.y - y.y, 2));
}

Point to(Point x, Point y, double a) {
  double xx, yy;
  if (x.x > y.x) {
    xx = x.x - (x.x - y.x) * a;
  } else {
    xx = x.x + (y.x - x.x) * a;
  }
  if (x.y > y.y) {
    yy = x.y - (x.y - y.y) * a;
  } else {
    yy = x.y + (y.y - x.y) * a;
  }
  return {xx, yy};
}

double Get() {
  double l = eps, r = 1;
  while (fabs(l - r) > eps) {
    double mid1 = l + (r - l) / 3, mid2 = r - (r - l) / 3;
    double ans1 = Dis(to(x[0], x[1], mid1), x[2]), ans2 = Dis(to(x[0], x[1], mid2), x[2]);
    if (ans2 > ans1) {
      r = mid2;
    } else {
      l = mid1;
    }
  }
  return l;
}

int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t;
  for (cin >> t; t; --t) {
    cin >> x[0].x >> x[0].y >> x[1].x >> x[1].y >> x[2].x >> x[2].y >> r;
    double res2 = max(Dis(x[0], x[2]), Dis(x[1], x[2])) + r;
    double ans = Get();
    cout << fixed << setprecision(2) << (Dis(to(x[0], x[1], ans), x[2]) - r) << ' ' << res2 << '\n';
  }
  cout << clock();
  return 0;
}

T2