Table of contents

# Problem

http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_2_D

# Solution

// C++ 14
#include <iostream>
#include <string>
#include <vector>
#include <list>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <unordered_map>
#include <math.h>

#define ll long long
#define Int int
#define loop(x, start, end) for(Int x = start; x < end; x++)
#define loopdown(x, start, end) for(int x = start; x > end; x--)
#define rep(n) for(int x = 0; x < n; x++)
#define span(a,x,y) a.begin()+x,a.begin()+y
#define span_all(a) a.begin(),a.end()
#define len(x) (x.size())
#define last(x) (*(x.end()-1))

using namespace std;

#define EPS 0.0000000001
#define fequals(a,b) (fabs((a) - (b)) < EPS)

class Vector2 {
public:
  double x, y;
  
  Vector2(double x = 0, double y = 0): x(x), y(y) {}
  
  Vector2 operator + (const Vector2 v) const { return Vector2(x + v.x, y + v.y); }
  Vector2 operator - (const Vector2 v) const { return Vector2(x - v.x, y - v.y); }
  Vector2 operator * (const double k) const { return Vector2(x * k, y * k); }
  Vector2 operator / (const double k) const { return Vector2(x / k, y / k); }
  
  double length() { return sqrt(norm()); }
  double norm() { return x * x + y * y; }
  double dot (Vector2 const v) { return x * v.x + y * v.y; }
  double cross (Vector2 const v) { return x * v.y - y * v.x; }
  
  bool parallel(Vector2 &other) {
    return fequals(0, cross(other));
  }
  
  bool orthogonal(Vector2 &other) {
    return fequals(0, dot(other));
  }
  
  bool operator < (const Vector2 &v) {
    return x != v.x ? x < v.x : y < v.y;
  }
  
  bool operator == (const Vector2 &v) {
    return fabs(x - v.x) < EPS && fabs(y - v.y) < EPS;
  }
};

ostream & operator << (ostream & out, Vector2 const & v) { 
  out<< "Vector2(" << v.x << ", " << v.y << ')';
  return out;
}

istream & operator >> (istream & in, Vector2 & v) { 
  double x, y;
  in >> x;
  in >> y;
  v.x = x;
  v.y = y;
  return in;
}

// Distance of a segment and a point
double p2Seg(Vector2 start_, Vector2 end_, Vector2 p) {
  if ((p - start_).dot(end_ - start_) < 0.0) return (p - start_).length();
  if ((p - end_).dot(start_ - end_) < 0.0) return (p - end_).length();
  return abs((end_ - start_).cross(p - start_)) / (end_ - start_).length();
}

double distance(Vector2 p1, Vector2 p2, Vector2 p3, Vector2 p4) {
  // Distance is 0 if 2 segments cross each other.
  double t3 = (p1.x - p2.x) * (p3.y - p1.y) + (p1.y - p2.y) * (p1.x - p3.x);
  double t4 = (p1.x - p2.x) * (p4.y - p1.y) + (p1.y - p2.y) * (p1.x - p4.x);
  double t1 = (p3.x - p4.x) * (p1.y - p3.y) + (p3.y - p4.y) * (p3.x - p1.x);
  double t2 = (p3.x - p4.x) * (p2.y - p3.y) + (p3.y - p4.y) * (p3.x - p2.x);
  if (t3 * t4 < 0 && t1 * t2 < 0) return 0.0;

  // Distance is the min of 4 distances of 4 points and 2 segments.
  double min_ = -1;
  min_ = min<double>(p2Seg(p1, p2, p3), p2Seg(p1, p2, p4));
  min_ = min<double>(min_, p2Seg(p3, p4, p1));
  min_ = min<double>(min_, p2Seg(p3, p4, p2));
  return min_;
}

#define MAX_N 1001
Int N;
Vector2 p1, p2, p3, p4;

void input() {
  cin >> N;
  while (cin >> p1 >> p2 >> p3 >> p4) {
    cout << distance(p1, p2, p3, p4) << endl;
  }
}

int main() {
  cout.precision(15);
  input();
}

CGL_2_D Distance

# Problem

# Solution

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