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 = 50000 + 10;
const double eps = 1e-6;
typedef long long LL;

void scan( int &x )
{
    char c;
    while( c = getchar(), c < '0' || c > '9' );
    x = c - '0';
    while( c = getchar(), c >= '0' && c <= '9' ) x = x*10 - c + '0';
}

struct Point
{
    double x, y;
    Point(double x=0, double y=0):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 = 1; i < n-1; ++i)
        ans += Cross(p[i]-p[0], p[i+1]-p[0]);
    return ans/2;
}

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) > 0;
}

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=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], q[last-1]);
    }
    while(first < last && !OnLeft(q[first], p[last-1])) last--;
    if(last-first <= 1) 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 = 0; i < n; ++i)
        lines.push_back(Line(p[(i+m+1)%n], p[i]-p[(i+m+1)%n]));
    return HalfplaneIntersection(lines).empty();
}

int solve()
{
    if(n==3) return 1;
    int L = 1, R = n-3, M;
    while(L < R)
    {
        M = L + (R-L)/2;
        if(check(M)) R = M;
        else L = M+1;
    }
    return L;
}

int main(void)
{
    #ifdef LOCAL
        freopen("4992in.txt", "r", stdin);
    #endif
    
    int x, y;
    while(scanf("%d", &n) == 1 && n)
    {
        for(int i = 0; i < n; ++i)
        {
            scanf("%d%d", &x, &y);
            p[i] = Point(x, y);
        }
        printf("%d
", solve());
    }
    
    return 0;
}