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)。

  1 //#define LOCAL
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <algorithm>
  5 #include <cmath>
  6 #include <vector>
  7 using namespace std;
  8 
  9 const int maxn = 50000 + 10;
 10 const double eps = 1e-6;
 11 typedef long long LL;
 12 
 13 void scan( int &x )
 14 {
 15     char c;
 16     while( c = getchar(), c < '0' || c > '9' );
 17     x = c - '0';
 18     while( c = getchar(), c >= '0' && c <= '9' ) x = x*10 - c + '0';
 19 }
 20 
 21 struct Point
 22 {
 23     double x, y;
 24     Point(double x=0, double y=0):x(x), y(y) {}
 25 };
 26 typedef Point Vector;
 27 Point operator + (const Point& a, const Point& b) { return Point(a.x+b.x, a.y+b.y); }
 28 Point operator - (const Point& a, const Point& b) { return Point(a.x-b.x, a.y-b.y); }
 29 Vector operator * (const Vector& a, double p) { return Vector(a.x*p, a.y*p); }
 30 double Dot(const Vector& a, const Vector& b) { return a.x*b.x + a.y*b.y; }
 31 double Cross(const Vector& a, const Vector& b) { return a.x*b.y - a.y*b.x; }
 32 double Length(const Vector& a) { return sqrt(Dot(a, a)); }
 33 Vector Normal(const Vector& a)
 34 {
 35     double l = Length(a);
 36     return Vector(-a.y/l, a.x/l);
 37 }
 38 
 39 double PolygonArea(const vector<Point>& p)
 40 {
 41     int n = p.size();
 42     double ans = 0.0;
 43     for(int i = 1; i < n-1; ++i)
 44         ans += Cross(p[i]-p[0], p[i+1]-p[0]);
 45     return ans/2;
 46 }
 47 
 48 struct Line
 49 {
 50     Point P;
 51     Vector v;
 52     double ang;
 53     Line() {}
 54     Line(Point p, Vector v):P(p), v(v) { ang = atan2(v.y, v.x); }
 55     bool operator < (const Line& L) const
 56     {
 57         return ang < L.ang;
 58     }
 59 };
 60 
 61 bool OnLeft(const Line& L, Point p)
 62 {
 63     return Cross(L.v, p-L.P) > 0;
 64 }
 65 
 66 Point GetLineIntersection(const Line& a, const Line& b)
 67 {
 68     Vector u = a.P - b.P;
 69     double t = Cross(b.v, u) / Cross(a.v, b.v);
 70     return a.P + a.v*t;
 71 }
 72 
 73 vector<Point> HalfplaneIntersection(vector<Line>& L)
 74 {
 75     int n = L.size();
 76     //sort(L.begin(), L.end());
 77     
 78     vector<Point> p(n);
 79     vector<Line> q(n);
 80     vector<Point> ans;
 81     int first, last;
 82     
 83     q[first=last=0] = L[0];
 84     for(int i = 1; i < n; ++i)
 85     {
 86         while(first < last && !OnLeft(L[i], p[last-1])) last--;
 87         while(first < last && !OnLeft(L[i], p[first])) first++;
 88         q[++last] = L[i];
 89         if(fabs(Cross(q[last].v, q[last-1].v)) < eps)
 90         {
 91             last--;
 92             if(OnLeft(q[last], L[i].P)) q[last] = L[i];
 93         }
 94         if(first < last) p[last-1] = GetLineIntersection(q[last], q[last-1]);
 95     }
 96     while(first < last && !OnLeft(q[first], p[last-1])) last--;
 97     if(last-first <= 1) return ans;
 98     p[last] = GetLineIntersection(q[first], q[last]);
 99     
100     for(int i = first; i <= last; ++i) ans.push_back(p[i]);
101     return ans;
102 }
103 
104 Point p[maxn];
105 int n;
106 
107 bool check(int m)
108 {
109     vector<Line> lines;
110     for(int i = 0; i < n; ++i)
111         lines.push_back(Line(p[(i+m+1)%n], p[i]-p[(i+m+1)%n]));
112     return HalfplaneIntersection(lines).empty();
113 }
114 
115 int solve()
116 {
117     if(n==3) return 1;
118     int L = 1, R = n-3, M;
119     while(L < R)
120     {
121         M = L + (R-L)/2;
122         if(check(M)) R = M;
123         else L = M+1;
124     }
125     return L;
126 }
127 
128 int main(void)
129 {
130     #ifdef LOCAL
131         freopen("4992in.txt", "r", stdin);
132     #endif
133     
134     int x, y;
135     while(scanf("%d", &n) == 1 && n)
136     {
137         for(int i = 0; i < n; ++i)
138         {
139             scanf("%d%d", &x, &y);
140             p[i] = Point(x, y);
141         }
142         printf("%d
", solve());
143     }
144     
145     return 0;
146 }

代码君