UVa 1475 (二分+半平面交) Jungle Outpost

题意：

有n个瞭望塔构成一个凸n边形，敌人会炸毁一些瞭望台，剩下的瞭望台构成新的凸包。在凸多边形内部选择一个点作为总部，使得敌人需要炸毁的瞭望塔最多才能使总部暴露出来。输出敌人需要炸毁的数目。

分析：

在炸毁同样数量的瞭望塔时，如何爆破才能使暴露出的面积最大。那就是集中火力炸掉连续的几个瞭望塔。直觉上是这样的，我不会证明这个结论。因为是连续爆破，所以k次爆破后还保留的部分就是一个半平面，枚举这k个爆破点，如果这些半平面交非空则总部可以设在这里。

k值是通过二分来确定的，下界是1，上界是n-3(这个需要好好想一下，想不清楚的话即使写n应该也可以)。

 //#define LOCAL
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <cmath>
 #include <vector>
 using namespace std;

 const int maxn =  + ;
 const double eps = 1e-;
 typedef long long LL;

 void scan( int &x )
 {
     char c;
     while( c = getchar(), c < '' || c > '' );
     x = c - '';
     while( c = getchar(), c >= '' && c <= '' ) x = x* - c + '';
 }

 struct Point
 {
     double x, y;
     Point(double x=, double y=):x(x), y(y) {}
 };
 typedef Point Vector;
 Point operator + (const Point& a, const Point& b) { return Point(a.x+b.x, a.y+b.y); }
 Point operator - (const Point& a, const Point& b) { return Point(a.x-b.x, a.y-b.y); }
 Vector operator * (const Vector& a, double p) { return Vector(a.x*p, a.y*p); }
 double Dot(const Vector& a, const Vector& b) { return a.x*b.x + a.y*b.y; }
 double Cross(const Vector& a, const Vector& b) { return a.x*b.y - a.y*b.x; }
 double Length(const Vector& a) { return sqrt(Dot(a, a)); }
 Vector Normal(const Vector& a)
 {
     double l = Length(a);
     return Vector(-a.y/l, a.x/l);
 }

 double PolygonArea(const vector<Point>& p)
 {
     int n = p.size();
     double ans = 0.0;
     for(int i = ; i < n-; ++i)
         ans += Cross(p[i]-p[], p[i+]-p[]);
     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;
     }
 };

 bool OnLeft(const Line& L, Point p)
 {
     return Cross(L.v, p-L.P) > ;
 }

 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;
 }

 vector<Point> HalfplaneIntersection(vector<Line>& L)
 {
     int n = L.size();
     //sort(L.begin(), L.end());

     vector<Point> p(n);
     vector<Line> q(n);
     vector<Point> ans;
     int first, last;

     q[first=last=] = L[];
     for(int i = ; i < n; ++i)
     {
         while(first < last && !OnLeft(L[i], p[last-])) last--;
         while(first < last && !OnLeft(L[i], p[first])) first++;
         q[++last] = L[i];
         if(fabs(Cross(q[last].v, q[last-].v)) < eps)
         {
             last--;
             if(OnLeft(q[last], L[i].P)) q[last] = L[i];
         }
         if(first < last) p[last-] = GetLineIntersection(q[last], q[last-]);
     }
     while(first < last && !OnLeft(q[first], p[last-])) last--;
     if(last-first <= ) return ans;
     p[last] = GetLineIntersection(q[first], q[last]);

     for(int i = first; i <= last; ++i) ans.push_back(p[i]);
     return ans;
 }

 Point p[maxn];
 int n;

 bool check(int m)
 {
     vector<Line> lines;
     for(int i = ; i < n; ++i)
         lines.push_back(Line(p[(i+m+)%n], p[i]-p[(i+m+)%n]));
     return HalfplaneIntersection(lines).empty();
 }

 int solve()
 {
     if(n==) return ;
     int L = , R = n-, M;
     while(L < R)
     {
         M = L + (R-L)/;
         if(check(M)) R = M;
         else L = M+;
     }
     return L;
 }

 int main(void)
 {
     #ifdef LOCAL
         freopen("4992in.txt", "r", stdin);
     #endif

     int x, y;
     while(scanf("%d", &n) ==  && n)
     {
         for(int i = ; i < n; ++i)
         {
             scanf("%d%d", &x, &y);
             p[i] = Point(x, y);
         }
         printf("%d\n", solve());
     }

     return ;
 }

代码君