Problem p02303 (C++) — buggy submission, verdict: Time Limit Exceeded

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <iostream>
using namespace std;
const int maxx = 1e5 + 6;

class point {
public:
  double x, y;
  point(double x = 0, double y = 0) : x(x), y(y) {}

  bool operator<(const point &u) const {
    if (x != u.x)
      return x < u.x;
    return y < u.y;
  }
};
point a[maxx], b[maxx];

inline double dis(point a, point b) {
  return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
inline bool cmp(point a, point b) { return a.y < b.y; }
double solve(int left, int right) {
  if (left + 1 == right)
    return dis(a[left], a[right]);
  double xl, xr, xm;
  int mid = (left + right) >> 1;
  xl = solve(left, mid);
  xr = solve(mid, right);
  xm = min(xl, xr);
  int k = 1;
  for (int i = left; i <= right; i++) {
    if (fabs(a[mid].x - a[i].x) <= xm)
      b[k++] = a[i];
  }
  sort(b + 1, b + k, cmp);
  for (int i = 1; i < k; i++) {
    for (int j = i + 1; j < k && (b[j].y - b[i].y < xm); j++) {
      xm = min(xm, dis(b[i], b[j]));
    }
  }
  return xm;
}

int main() {
  int n;
  double x, y;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lf%lf", &x, &y);
    a[i] = point(x, y);
  }
  sort(a + 1, a + 1 + n);
  double ans = solve(1, n);
  printf("%.6f\\n", ans);
  return 0;
}


Error: TLE

Fixed submission:

#include <bits/stdc++.h>

using namespace std;

#define EPS (1e-10)
#define equals(a, b) (fabs((a) - (b)) < EPS)

const double inf = 1e20;

inline int sign(const double &x) {
  if (x > EPS)
    return 1;
  else if (x < -EPS)
    return -1;
  return 0;
}

/*   点  */
struct Point {
  double x, y;

  Point(double x = 0, double y = 0) : x(x), y(y) {}

  Point operator+(Point p) { return Point(x + p.x, y + p.y); }
  Point operator-(Point p) { return Point(x - p.x, y - p.y); }
  Point operator*(double k) { return Point(k * x, k * y); }
  Point operator/(double k) { return Point(x / k, y / k); }

  // 向量的大小
  double abs() { return sqrt(norm()); }
  double norm() { return x * x + y * y; }

  bool operator<(const Point &p) const { return x != p.x ? x < p.x : y < p.y; }

  bool operator==(const Point &p) const {
    return fabs(x - p.x) < EPS && fabs(y - p.y) < EPS;
  }
};

// 两点间的距离
double getDistance(Point a, Point b) { return (a - b).abs(); }

/*   向量  */
typedef Point Vector;

// 向量a和b的内积
double dot(Vector a, Vector b) { return a.x * b.x + a.y * b.y; }
// 向量a和b的外积
double cross(Vector a, Vector b) { return a.x * b.y - a.y * b.x; }

/*   线段  */
struct Segment {
  Point p1, p2;
};

// 点p在线段s上的投影
Point project(Segment s, Point p) {
  Vector base = s.p2 - s.p1;
  double r = dot(p - s.p1, base) / base.norm();
  return s.p1 + base * r;
}

Point reflect(Segment s, Point p) { return p + (project(s, p) - p) * 2.0; }

/*   直线  */
typedef Segment Line;

// 点到直线的距离
double getDistanceLP(Line l, Point p) {
  return abs(cross(l.p2 - l.p1, p - l.p1) / (l.p2 - l.p1).abs());
}

// 点到线段的距离
double getDistanceSP(Segment s, Point p) {
  if (dot(s.p2 - s.p1, p - s.p1) < 0.0)
    return (p - s.p1).abs();
  if (dot(s.p1 - s.p2, p - s.p2) < 0.0)
    return (p - s.p2).abs();
  return getDistanceLP(s, p);
}

// 判断p2与向量p1-p0的位置关系

const int COUNTER_CLOCKWISE = -1; // 逆时针
const int CLOCKWISE = 1;          // 顺时针
const int ON_SEGMENT = 0;         // 在线段上
const int ONLINE_BACK = 2;        // p2在向量p1-p0的反方向延长线上
const int ONLINE_FRONT = -2;      // p2在向量p1-p0的正方向延长线上

int ccw(Point p0, Point p1, Point p2) {
  Vector a = p1 - p0;
  Vector b = p2 - p0;
  if (cross(a, b) > EPS)
    return COUNTER_CLOCKWISE;
  if (cross(a, b) < -EPS)
    return CLOCKWISE;
  if (dot(a, b) < -EPS)
    return ONLINE_BACK;
  if (a.norm() < b.norm())
    return ONLINE_FRONT;
  return ON_SEGMENT;
}

// 判断线段p1p2与线段p3p4是否相交
bool intersect(Point p1, Point p2, Point p3, Point p4) {
  return (ccw(p1, p2, p3) * ccw(p1, p2, p4) <= 0 &&
          ccw(p3, p4, p1) * ccw(p3, p4, p2) <= 0);
}

bool intersect(Segment s1, Segment s2) {
  return intersect(s1.p1, s1.p2, s2.p1, s2.p2);
}

// 线段与线段的距离
double getDistance(Segment s1, Segment s2) {
  if (intersect(s1, s2))
    return 0.0;
  return min(min(getDistanceSP(s1, s2.p1), getDistanceSP(s1, s2.p2)),
             min(getDistanceSP(s2, s1.p1), getDistanceSP(s2, s1.p2)));
}

// 两个线段的交点(对于直线跟线段也适用，但注意base是在直线上)
Point getCrossPoint(Segment s1, Segment s2) {
  Vector base = s2.p2 - s2.p1;
  double d1 = abs(cross(base, s1.p1 - s2.p1));
  double d2 = abs(cross(base, s1.p2 - s2.p1));
  double t = d1 / (d1 + d2);
  return s1.p1 + (s1.p2 - s1.p1) * t;
}

/* 直线正交
 * 判断向量a，b是否正交 =》a，b的内积为0
 */
bool isOrthogonal(Vector a, Vector b) { return equals(dot(a, b), 0.0); }

bool isOrthogonal(Point a1, Point a2, Point b1, Point b2) {
  return isOrthogonal(a1 - a2, b1 - b2);
}

/* 直线平行
 * 判断向量a，b是否平行 =》a，b的内积为0
 */
bool isParallel(Vector a, Vector b) { return equals(cross(a, b), 0.0); }

bool isParallel(Point a1, Point a2, Point b1, Point b2) {
  return isParallel(a1 - a2, b1 - b2);
}

struct Circle {
  Point c;
  double r;
  Circle(Point c = Point(), double r = 0.0) : c(c), r(r) {}
};

/* 圆与直线的交点
 * 交点为一个的时候返回两个相同的点
 * 没有交点则退出
 */

bool intersect(Circle c, Line l) {
  return sign(getDistanceLP(l, c.c) - c.r) <= 0;
}

pair<Point, Point> getCrossPoints(Circle c, Line l) {
  assert(intersect(c, l)); // 没有交点
  Point pr = project(l, c.c);
  Vector e = (l.p2 - l.p1) / (l.p2 - l.p1).abs();
  double base = sqrt(c.r * c.r - (pr - c.c).norm());
  return make_pair(pr + e * base, pr - e * base);
}

// 圆与圆的交点

bool intersect(Circle c1, Circle c2) {
  return sign(getDistance(c1.c, c2.c) - (c1.r + c2.r)) <= 0;
}

double arg(Vector p) {
  return atan2(p.y, p.x); // double atan2(double y,double x)
                          // 返回的是原点至点(x,y)的方位角，即与 x
                          // 轴的夹角。返回值的单位为弧度，取值范围为(-PI,PI]
}

Vector polar(double r, double a) { return Point(cos(a) * r, sin(a) * r); }

pair<Point, Point> getCrossPoints(Circle c1, Circle c2) {
  assert(intersect(c1, c2));
  double d = (c1.c - c2.c).abs();
  double a = acos((c1.r * c1.r + d * d - c2.r * c2.r) / (2 * c1.r * d));
  double t = arg(c2.c - c1.c);
  return make_pair(c1.c + polar(c1.r, t + a), c1.c + polar(c1.r, t - a));
}

/*   多边形  */
typedef vector<Point> Polygon;

// 点的****内包*****(判断点与多边形的关系)
// 多边形内返回2
// 多边形上返回1
// 多边形外返回0
int contains(Polygon g, Point p) {
  int n = g.size();
  bool x = false;
  for (int i = 0; i < n; i++) {
    Point a = g[i] - p, b = g[(i + 1) % n] - p;
    if (abs(cross(a, b)) < EPS && dot(a, b) < EPS)
      return 1;
    if (a.y > b.y)
      swap(a, b);
    if (a.y < EPS && EPS < b.y && cross(a, b) > EPS)
      x = !x;
  }
  return (x ? 2 : 0);
}

/* *******凸包*****(包含点集合p中所有点的最小凸多边形)
 * 输出凸多边形最下端最左侧的顶点为起点,按逆时针方向依次输出坐标。
 * 安德鲁算法 O(n*logn)
 */

double Area(Point p0, Point p1, Point p2) {
  return (p1.x - p0.x) * (p2.y - p0.y) - (p2.x - p0.x) * (p1.y - p0.y);
}

Polygon andrewscanf(Polygon s) {
  int n = s.size();
  sort(s.begin(), s.end()); // x从小到大，x相同的y从小到大
  Polygon u;
  // 构建凸包上部
  for (int i = 0; i < n; i++) {
    for (int j = u.size(); j >= 2 && Area(u[j - 2], u[j - 1], s[i]) <= EPS;
         j--) {
      u.pop_back();
    }
    u.push_back(s[i]);
  }
  // 构建凸包下部
  int tmp = u.size() + 1;
  for (int i = n - 2; i >= 0; i--) {
    for (int j = u.size(); j >= tmp && Area(u[j - 2], u[j - 1], s[i]) <= EPS;
         j--) {
      u.pop_back();
    }
    u.push_back(s[i]);
  }
  // 按顺序生成凸包点的序列
  if (n > 1)
    u.pop_back(); // 开始点多加入了一次
  return u;
}

// 旋转卡壳*-------》计算凸包直径
double RC(Polygon v) {
  double res = 0;
  int n = v.size();
  int k = 1;
  for (int i = 0; i < n; i++) {
    while (sign(Area(v[i], v[(i + 1) % n], v[(k + 1) % n]) -
                Area(v[i], v[(i + 1) % n], v[k % n])) == 1)
      k = (k + 1) % n;
    //        printf("*** %d\\n",k);
    res = max(res, max((v[i] - v[k]).abs(), (v[(i + 1) % n] - v[k]).abs()));
  }
  return res;
}

//  平面上最近点对。（分治  nlognlogn）

inline bool Closest_Pair_CmpY(Point a, Point b) { return a.y < b.y; }

double Closest_Pair_Divide_Conquer(int l, int r, Polygon g, int n) {
  if (l == r)
    return inf;
  if (l + 1 == r)
    return getDistance(g[l], g[r]);
  int mid = l + r >> 1;
  double ans = min(Closest_Pair_Divide_Conquer(l, mid, g, n),
                   Closest_Pair_Divide_Conquer(mid + 1, r, g, n));
  Polygon temp(n); // 保存中点左右满足条件的点
  int len = 0;
  for (int i = l; i <= r; i++) {
    if (fabs(g[mid].x - g[i].x) <= ans)
      temp[len++] = g[i];
  }
  sort(temp.begin(), temp.begin() + len, Closest_Pair_CmpY);
  for (int i = 0; i < len; i++) {
    for (int j = i + 1; j < len && (temp[j].y - temp[i].y < ans); j++) {
      ans = min(ans, getDistance(temp[i], temp[j]));
    }
  }
  return ans;
}

inline void Closest_Pair_Solve(Polygon g, int n) {
  sort(g.begin(), g.end());
  printf("%.10f\\n", Closest_Pair_Divide_Conquer(0, n - 1, g, n));
}

int main() {
  int n;
  Point a;
  scanf("%d", &n);
  Polygon g(n);
  for (int i = 0; i < n; i++) {
    scanf("%lf%lf", &g[i].x, &g[i].y);
  }
  Closest_Pair_Solve(g, n);
  return 0;
}