HDU-3644 A Chocolate Manufacturer’s Problem

題意: 給定一個多邊形, 判斷這個多邊形中是否可以放入一個半徑為r的圓.
分析: 發現不知從何入手時就開始模擬退火吧. 隨機找出圓心座標, 主要就是判斷某個點是否在多邊形內. 這題wa和tle了好多次, 引數選擇需要些微調, 模擬退火有風險, 罰時傷不起.
程式碼:

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <ctime>
 

#include <iostream>

using namespace std;

const int MAXN = 111;
const double pi = acos(-1);
const double inf = 0x3f3f3f3f;
const double eps = 1e-4;

int sgn(double x) {
  if (fabs(x) < eps)
    return 0;
  else if (x < 0)
    return -1;
  else
    return 1;
}

struct Point {
  double x, y;
 

  Point() {}
  Point(double _x, double _y) {
    x = _x;
    y = _y;
  }

  Point operator+(const Point b) const { return Point(x + b.x, y + b.y); }
  Point operator-(const Point b) const { return Point(x - b.x, y - b.y); }
  double operator*(const Point b) const { return x * b.x + y * b.y; }
  double
 
 operator^(const Point b) const { return x * b.y - y * b.x; }
  bool operator==(const Point b) {
    return sgn(x - b.x) == 0 && sgn(y - b.y) == 0;
  }
  double distance(Point b) { return hypot(x - b.x, y - b.y); }
};

struct Line {
  Point s, e;
  Line() {}
  Line(Point _s, Point _e) {
    s = _s;
    e = _e;
  }

  double length() { return s.distance(e); }

  double dispointtoline(Point b) { return fabs((b - s) ^ (e - s)) / length(); }
  double dispointtoseg(Point b) {
    if (sgn((b - s) * (e - s)) < 0 || sgn((b - e) * (s - e)) < 0)
      return min(b.distance(s), b.distance(e));
    else
      return dispointtoline(b);
  }
  bool pointonseg(Point b) {
    return sgn((b - s) ^ (e - s)) == 0 && sgn((b - s) * (b - e)) <= 0;
  }
};

struct Polygon {
  Point p[MAXN];
  Line l[MAXN];
  int n;
  void add(Point b) { p[n++] = b; }
  void getline() {
    for (int i = 0; i < n; i++) {
      l[i] = Line(p[i], p[(i + 1) % n]);
    }
  }

  int relationpoint(Point q) {
    for (int i = 0; i < n; i++) {
      if (p[i] == q)
        return 3;
    }
    getline();
    for (int i = 0; i < n; i++) {
      if (l[i].pointonseg(q))
        return 2;
    }
    int cnt = 0;
    for (int i = 0; i < n; i++) {
      int j = (i + 1) % n;
      int k = sgn((q - p[i]) ^ (p[i] - p[j]));
      int u = sgn(p[i].y - q.y);
      int v = sgn(p[j].y - q.y);
      if (k > 0 && u < 0 && v >= 0)
        cnt++;
      if (k < 0 && v < 0 && u >= 0)
        cnt--;
    }
    return cnt != 0;
  }

  double getdis(Point cir) {
    double res = inf;
    getline();
    for (int i = 0; i < n; i++) {
      res = min(res, l[i].dispointtoseg(cir));
    }
    // cout << res << endl;
    return res;
  }
};

Polygon pol;
int n;
double r;
Point ans[MAXN];

double Rand() { return (double)rand() / RAND_MAX; }
int main() {
  srand(time(NULL));
  while (scanf("%d", &n) != EOF) {
    if (n == 0)
      break;
    pol.n = 0;
    for (int i = 0; i < n; i++) {
      double x, y;
      scanf("%lf%lf", &x, &y);
      pol.add(Point(x, y));
    }
    scanf("%lf", &r);
    double x1 = pol.p[0].x, x2 = pol.p[0].x, y1 = pol.p[0].y, y2 = pol.p[0].y;
    for (int i = 0; i < n; i++) {
      ans[i].x = (pol.p[i].x + pol.p[(i + 1) % n].x) / 2.0;
      ans[i].y = (pol.p[i].y + pol.p[(i + 1) % n].y) / 2.0;
      x1 = min(x1, pol.p[i].x);
      x2 = max(x2, pol.p[i].x);
      y1 = min(y1, pol.p[i].y);
      y2 = max(y2, pol.p[i].y);
    }
    double res = -inf;
    double T = sqrt((x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1)) / 2;
    // cout << " --------------- " << endl;
    bool flag = 0;
    while (T > eps && !flag) {
      for (int j = 0; j < n; j++) {
        for (int i = 0; i < 5; i++) {
          Point nxt;
          nxt.x = ans[j].x + T * (Rand() * 2 - 1);
          nxt.y = ans[j].y + T * (Rand() * 2 - 1);

          if (pol.relationpoint(nxt) != 1)
            continue;
          // cout << nxt.x << " " << nxt.y << endl;

          double dis = pol.getdis(nxt);
          if (sgn(res - dis) < 0) {
            res = dis;
            ans[j] = nxt;
            if (sgn(res - r) >= 0)
              flag = 1;
          }
        }
      }
      T *= 0.9;
    }
    // cout << res << endl;
    printf("%s\n", !flag ? "No" : "Yes");
  }
  return 0;
}