uvalive 4992(半平面交+二分)

题意：丛林里有n个瞭望塔，组成了一个凸多边形，瞭望塔可以保护这个凸多边形内的任意一点，总部就设在凸多边形内的某一点，敌人会炸掉瞭望塔使总部不再被剩下的塔监视到，现在要选择一个最优的总部位置，输出敌人需要炸毁的瞭望塔的数目。

题解：二分出需要炸的数目然后把剩下的塔拿去半平面交看是否还有交集，可以想到如果有两颗炸弹，敌人会选择连续的两个顶点炸，因为这样损失的面积更大，然后我们也这样考虑半平面交时候的各边。注意逆向输入，因为半平面取的是边的左边而题目是顺时针输入。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;
const double PI = acos(-1);
const int N = 50010;
const double INF = 1e9;
const double eps = 1e-7;
struct Point {
double x, y;
Point(double x = 0, double y = 0): x(x), y(y) {}
}P
;

typedef Point Vector;

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;
}
};
int n;

double Sqr(double x) {
return x * x;
}
Point operator + (Point A, Point B) {
return Point(A.x + B.x, A.y + B.y);
}
Point operator - (Point A, Point B) {
return Point(A.x - B.x, A.y - B.y);
}
Point operator * (Point A, double p) {
return Point(A.x * p, A.y * p);
}
Point operator / (Point A, double p) {
return Point(A.x / p, A.y / p);
}
//计算点积的正负  负值夹角为钝角
int dcmp(double x) {
if (fabs(x) < 1e-9)
return 0;
return x < 0 ? -1 : 1;
}
bool operator < (const Point& a, const Point& b) {
return a.x < b.x || (a.x == b.x && a.y < b.y);
}
bool operator == (const Point& a, const Point& b) {
return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}
//计算点积
double Dot(Point A, Point B) {
return A.x * B.x + A.y * B.y;
}
//计算叉积，也就是数量积
double Cross(Point A, Point B) {
return A.x * B.y - A.y * B.x;
}
//计算向量长度
double Length(Point A) {
return sqrt(Dot(A, A));
}
Vector Normal(Vector A) {
double L = Length(A);
return Vector(-A.y / L, A.x / L);
}
//向量A旋转rad弧度，rad负值为顺时针旋转
Vector Rotate(Vector A, double rad) {
return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));
}
//角度转化弧度
double torad(double deg) {
return deg / 180.0 * PI;
}
//点p在有向直线L的左边(线上不算)
bool OnLeft(Line L, Point P) {
return Cross(L.v, P - L.p) > 0;
}
//求两直线的交点，前提交点一定存在
Point GetIntersection(Line a, 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;
}
//求半面交
int HalfplaneIntersection(vector<Line>& L) {
vector<Point> poly;
int n = L.size();
sort(L.begin(), L.end());
int first = 0, rear = 0;
vector<Point> p(n);
vector<Line> q(n);

q[first] = L[0];
for (int i = 1; i < n; i++) {
while (first < rear && !OnLeft(L[i], p[rear - 1]))
rear--;
while (first < rear && !OnLeft(L[i], p[first]))
first++;
q[++rear] = L[i];
if (fabs(Cross(q[rear].v, q[rear - 1].v)) < eps) {
rear--;
if (OnLeft(q[rear], L[i].p))
q[rear] = L[i];
}
if (first < rear)
p[rear - 1] = GetIntersection(q[rear - 1], q[rear]);
}
while (first < rear && !OnLeft(q[first], p[rear - 1]))
rear--;
if (rear - first <= 1)
return 0;
p[rear] = GetIntersection(q[rear], q[first]);
for (int i = first; i <= rear; i++)
poly.push_back(p[i]);
return poly.size();
}

int main() {
while (scanf("%d", &n) == 1 && n) {
for (int i = 0; i < n; i++)
scanf("%lf%lf", &P[i].x, &P[i].y);
int left = 0, right = n;
while (left < right) {
vector<Line> L;
int mid = (left + right) / 2;
for (int i = 0; i < n; i++)
L.push_back(Line( P[(i + 1 + mid) % n], P[i] - P[(i + 1 + mid) % n]));
if (!HalfplaneIntersection(L))
right = mid;
else
left = mid + 1;
}
printf("%d\n", left);
}
return 0;
}