Jungle Outpost UVALive - 4992（贪心+二分+半平面交）

题目链接：https://cn.vjudge.net/problem/UVALive-4992

题意：有n个瞭望台，形成一个凸多边形，这些瞭望台的保护范围就是凸边形内的任意一点，敌人进攻时，会炸毁一些了瞭望台，

使得总部暴露在那些剩下的瞭望台的凸包之外，要求选择一个点作为总部，是的敌人需要炸毁的瞭望台数量尽可能的多，

题解：这是一个最小值最大化的问题，考虑敌人如何能用最少的炸弹，使得剩下的瞭望台形成的凸包的面积最小？

不难发现，那必然是炸连续的瞭望台，纸上画一画，想一想。

想到这里就很简单了，我们可以二分答案，假如答案是m，然后check所有每m个连续连续顶点对应的新半平面的交集是否为空集，如果交集不为空，那么如果总部建在交集内的任意一点，那么无论从哪个点出发，删除任意的连续的m个点都不能使得总部暴露在凸包之外。

代码：

# define _CRT_SECURE_NO_WARNINGS
//#pragma GCC optimize(2)
//#include <bits/stdc++.h>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <vector>
#define inf 0x3f3f3f3f
#define Pair pair<int,int>
//#define int long long
#define fir first
#define sec second
namespace fastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
  //fread->read
  bool IOerror = 0;
  //inline char nc(){char ch=getchar();if(ch==-1)IOerror=1;return ch;}
  inline char nc() {
    static char buf[BUF_SIZE], * p1 = buf + BUF_SIZE, * pend = buf + BUF_SIZE;
    if (p1 == pend) {
      p1 = buf; pend = buf + fread(buf, 1, BUF_SIZE, stdin);
      if (pend == p1) { IOerror = 1; return -1; }
    }
    return *p1++;
  }
  inline bool blank(char ch) { return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t'; }
  template<class T> inline bool read(T& x) {
    bool sign = 0; char ch = nc(); x = 0;
    for (; blank(ch); ch = nc());
    if (IOerror)return false;
    if (ch == '-')sign = 1, ch = nc();
    for (; ch >= '0' && ch <= '9'; ch = nc())x = x * 10 + ch - '0';
    if (sign)x = -x;
    return true;
  }
  inline bool read(double& x) {
    bool sign = 0; char ch = nc(); x = 0;
    for (; blank(ch); ch = nc());
    if (IOerror)return false;
    if (ch == '-')sign = 1, ch = nc();
    for (; ch >= '0' && ch <= '9'; ch = nc())x = x * 10 + ch - '0';
    if (ch == '.') { double tmp = 1; ch = nc(); for (; ch >= '0' && ch <= '9'; ch = nc())tmp /= 10.0, x += tmp * (ch - '0'); }
    if (sign)x = -x; return true;
  }
  inline bool read(char* s) {
    char ch = nc();
    for (; blank(ch); ch = nc());
    if (IOerror)return false;
    for (; !blank(ch) && !IOerror; ch = nc())* s++ = ch;
    *s = 0;
    return true;
  }
  inline bool read(char& c) {
    for (c = nc(); blank(c); c = nc());
    if (IOerror) { c = -1; return false; }
    return true;
  }
  template<class T, class... U>bool read(T& h, U& ... t) { return read(h) && read(t...); }
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace fastIO;
using namespace std;

const int N = 2e5 + 5;
const double eps = 1e-10;
const double pi = acos(-1.0);
const int mod = 998244353;
/*---------------------------------------------------------------------------------------------------------------------------*/
struct Point { double x, y; Point(double x = 0, double y = 0) :x(x), y(y) {} };
typedef Point Vector;
Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }
Vector operator - (Point A, Point B) { return Vector(A.x - B.x, A.y - B.y); }
Vector operator * (Vector A, double p) { return Vector(A.x * p, A.y * p); }
Vector operator / (Vector A, double p) { return Vector(A.x / p, A.y / p); }
bool operator < (const Point& A, const Point& B) { return A.x < B.x || (A.x == B.x && A.y < B.y); }
int dcmp(double x) { if (fabs(x) < eps)return 0; else return x < 0 ? -1 : 1; }
bool operator == (const Point& A, const Point& B) { return dcmp(A.x - B.x) == 0 && dcmp(A.y - B.y) == 0; }
double Dot(Vector A, Vector B) { return A.x * B.x + A.y * B.y; }
double Length(Vector A) { return sqrt(Dot(A, A)); }
double Dist2(const Point& A, const Point& B) { return (A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y); }
double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }
double Cross(Vector A, Vector B) { return A.x * B.y - A.y * B.x; }
double Area2(Point A, Point B, Point C) { return Cross(B - A, C - A); }
double DistanceToLine(Point P, Point A, Point B) { Vector v1 = B - A, v2 = P - A; return fabs(Cross(v1, v2)) / Length(v1); }
Vector Normal(Vector A) { double L = Length(A); return Vector(-A.y / L, A.x / L); }
Vector Rotate(Vector A, double rad) { return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad)); }
Point GetLineIntersection(Point P, Vector v, Point Q, Vector w) { Vector u = P - Q; double t = Cross(w, u) / Cross(v, w); return P + v * t; }
double PolygonArea(vector<Point>p) {//duo area
  double area = 0;
  int n = p.size();
  for (int i = 1; i < n - 1; i++)area += Cross(p[i] - p[0], p[i + 1] - p[0]);
  return area / 2;
}
double torad(double deg) { return deg / 180 * acos(-1); }
vector<Point> ConvexHull(vector<Point>& p) {
  sort(p.begin(), p.end());
  p.erase(unique(p.begin(), p.end()), p.end());
  int n = p.size();
  int m = 0;
  vector<Point> ch(n + 1);
  for (int i = 0; i < n; i++) { while (m > 1 && Cross(ch[m - 1] - ch[m - 2], p[i] - ch[m - 2]) <= 0)m--; ch[m++] = p[i]; }
  int k = m;
  for (int i = n - 2; i >= 0; i--) { while (m > k && Cross(ch[m - 1] - ch[m - 2], p[i] - ch[m - 2]) <= 0)m--; ch[m++] = p[i]; }
  if (n > 1)m--;
  ch.resize(m);
  return ch;
}
double diameter2(vector<Point>& points, Point& A, Point& B) {
  vector<Point>p = ConvexHull(points);
  int n = p.size();
  if (n == 1) return 0;
  if (n == 2) return Dist2(p[0], p[1]);
  //p.push_back(p[0]); // 免得取模
  double ans = 0;
  for (int u = 0, v = 1; u < n; u++) {
    for (;;) {
      //Cross(p[u+1]-p[u],p[v+1]-p[v])<=0时停止旋转
      int diff = Cross(p[u + 1] - p[u], p[v + 1] - p[v]);
      if (diff <= 0) {
        if (Dist2(p[u], p[v]) > ans) {
          ans = Dist2(p[u], p[v]); // u和v是对踵点
          A = p[u]; B = p[v];
        }
        //if(diff==0) ans=max(ans,Dist2(p[u],p[v+1])); // diff==0 时u和v+1也是对踵点
        break;
      }
      v = (v + 1) % n;
    }
  }
  return ans;
}
/*---------------------------------------------------------------------------------------------------------------------------*/
struct Line {
  Point P;
  Vector v;
  double ang;
  Line() {}
  Line(Point P, Vector v) :P(P), v(v) { ang = atan2(v.y, v.x); }
  bool operator < (const Line& L)const {
    return ang < L.ang;
  }
};
Point GetLineIntersection(const Line& a, const Line& b) {
  Vector u = a.P - b.P;
  double t = Cross(b.v, u) / Cross(a.v, b.v);
  return a.P + a.v * t;
}
bool OnLeft(Line L, Point p) { return Cross(L.v, p - L.P) > 0; }
vector<Point>HalfplaneIntersection(vector<Line>L) {
  int n = L.size();
  sort(L.begin(), L.end());
  int first, last;
  vector<Point> p(n);
  vector<Line> q(n);
  vector<Point> ans;
  q[first = last = 0] = L[0];
  for (int i = 1; i < n; i++) {
    while (first < last && !OnLeft(L[i], p[last - 1])) last--;
    while (first < last && !OnLeft(L[i], p[first]))first++;
    q[++last] = L[i];
    if (fabs(Cross(q[last].v, q[last - 1].v)) < eps) {
      last--;
      if (OnLeft(q[last], L[i].P))q[last] = L[i];
    }
    if (first < last)p[last - 1] = GetLineIntersection(q[last - 1], q[last]);
  }
  while (first < last && !OnLeft(q[first], p[last - 1]))last--;//删除无用平面
  if (last - first <= 1)return ans;
  p[last] = GetLineIntersection(q[last], q[first]);//计算首尾两个半平面的交点
  //从deque复制到输出中
  for (int i = first; i <= last; i++)ans.push_back(p[i]);
  return ans;
}
bool check(int m, vector<Point>P) {
  vector<Line>L;
  int n = P.size();
  for (int i = 0; i < n; i++)
    L.push_back(Line(P[(i + 1 + m) % n], P[i] - P[(i + 1 + m) % n]));
  return HalfplaneIntersection(L).empty();
}
/*---------------------------------------------------------------------------------------------------------------------------*/

signed main() {
  int n;
  while (scanf("%d", &n) == 1 && n) {
    vector<Point>P;
    for (int i = 0; i < n; i++) {
      double x, y;
      scanf("%lf%lf", &x, &y);
      P.push_back(Point(x, y));
    }
    if (n == 3) {
      printf("1\n");
      continue;
    }
    //最小值最大化
    int l = 1, r = n - 3;
    while (l < r) {
      int mid = l + r >> 1;
      //int mid = l + r >> 1;
      if (check(mid, P)) r = mid;
      else l = mid + 1;
    }
    printf("%d\n", l);
  }
  return 0;
}