HDU 3761 (二分 半平面交)

题目链接：点击这里

题意：顺时针给出一个凸包， 问最少炸掉多少点使得凸包面积为0.

炸掉的点必然是连续的。 然后二分一下结果， 每次判断半平面的面积。 注意chenk的时候要check面积而不是半平面交的顶点个数。 注意顺时针输入这个trick。

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;
#define maxn 100005

const double eps = 1e-8;
const double INF = 1e20;
const double pi = acos (-1.0);

int dcmp (double x) {
    if (fabs (x) < eps) return 0;
    return (x < 0 ? -1 : 1);
}
inline double sqr (double x) {return x*x;}

//*************点
struct Point {
    double x, y;
    Point (double _x = 0, double _y = 0):x(_x), y(_y) {}
    void input () {scanf ("%lf%lf", &x, &y);}
    void output () {printf ("%.2f %.2f\n", x, y);}
    bool operator == (const Point &b) const {
        return (dcmp (x-b.x) == 0 && dcmp (y-b.y) == 0);
    }
    bool operator < (const Point &b) const {
        return (dcmp (x-b.x) == 0 ? dcmp (y-b.y) < 0 : x < b.x);
    }
    Point operator + (const Point &b) const {
        return Point (x+b.x, y+b.y);
    }
    Point operator - (const Point &b) const {
        return Point (x-b.x, y-b.y);
    }
    Point operator * (double a) {
        return Point (x*a, y*a);
    }
    Point operator / (double a) {
        return Point (x/a, y/a);
    }
    double len2 () {//返回长度的平方
        return sqr (x) + sqr (y);
    }
    double len () {//返回长度
        return sqrt (len2 ());
    }
    Point change_len (double r) {//转化为长度为r的向量
        double l = len ();
        if (dcmp (l) == 0) return *this;//零向量返回自身
        r /= l;
        return Point (x*r, y*r);
    }
    Point rotate_left () {//顺时针旋转90度
        return Point (-y, x);
    }
    Point rotate_right () {//逆时针旋转90度
        return Point (y, -x);
    }
    Point rotate (Point p, double ang) {//绕点p逆时针旋转ang
        Point v = (*this)-p;
        double c = cos (ang), s = sin (ang);
        return Point (p.x + v.x*c - v.y*s, p.y + v.x*s + v.y*c);
    }
    Point normal () {//单位法向量
        double l = len ();
        return Point (-y/l, x/l);
    }
};

double cross (Point a, Point b) {//叉积

    return a.x*b.y-a.y*b.x;
}
double dot (Point a, Point b) {//点积

    return a.x*b.x + a.y*b.y;
}
double dis (Point a, Point b) {//两个点的距离

    Point p = b-a; return p.len ();
}
double rad_degree (double rad) {//弧度转化为角度

    return rad/pi*180;
}
double rad (Point a, Point b) {//两个向量的夹角

    return fabs (atan2 (fabs (cross (a, b)), dot (a, b)) );
}
bool parallel (Point a, Point b) {//向量平行
    double p = rad (a, b);
    return dcmp (p) == 0 || dcmp (p-pi) == 0;
}

//************直线 线段
struct Line {
    Point s, e;//直线的两个点
    double k;//极角
    Line () {}
    Line (Point _s, Point _e) {
        s = _s, e = _e;
        k = atan2 (e.y - s.y,e.x - s.x);
    }
    //一个点和倾斜角确定直线
    Line (Point p, double ang) {
        k = ang;
        s = p;
        if (dcmp (ang-pi/2) == 0) {
            e = s + Point (0, 1);
        }
        else
            e = s + Point (1, tan (ang));
    }
    void input () {
        s.input ();
        e.input ();
    }
    void adjust () {
        if (e < s) swap (e, s);
    }
    double length () {//求线段长度
        return dis (s, e);
    }
    void get_angle () {
        k = atan2 (e.y - s.y,e.x - s.x);
    }
    double angle () {//直线的倾斜角
        if (dcmp (k) < 0) k += pi;
        if (dcmp (k-pi) == 0) k -= pi;
        return k;
    }
    Point operator &(const Line &b)const {//直线的交点(保证存在)
        Point res = s;
        double t = (cross (s - b.s, b.s - b.e))/cross (s - e, b.s - b.e); 
        res.x += (e.x - s.x)*t;
        res.y += (e.y - s.y)*t;
        return res;
    }
};

double polygon_area (Point *p, int n) {
    //n个点
    double area = 0;
    for (int i = 1; i < n-1; i++) {
        area += cross (p[i]-p[0], p[i+1]-p[0]);
    }
    return area/2;
}

bool HPIcmp (const Line &a, const Line &b) {
    if (fabs(a.k - b.k) > eps)
        return a.k < b.k;
    return cross (a.s - b.s, b.e - b.s) < 0; 
}
Line Q[maxn];
void HPI(Line line[], int n, Point res[], int &resn) {
    int tot = n; 
    sort(line,line+n,HPIcmp); 
    tot = 1;
    for(int i = 1;i < n;i++){
        if(fabs(line[i].k - line[i-1].k) > eps)
            line[tot++] = line[i];
    }
    int head = 0, tail = 1; 
    Q[0] = line[0];
    Q[1] = line[1];
    resn = 0;
    for (int i = 2; i < tot; i++) {
        if (fabs(cross (Q[tail].e-Q[tail].s, Q[tail-1].e-Q[tail-1].s)) < eps || fabs(cross (Q[head].e-Q[head].s, Q[head+1].e-Q[head+1].s)) < eps)
            return;
        while(head < tail && (cross ((Q[tail]&Q[tail-1])-line[i].s, line[i].e-line[i].s)) > eps) tail--;
        while(head < tail && (cross ((Q[head]&Q[head+1]) - line[i].s, line[i].e-line[i].s)) > eps)
        head++; 
        Q[++tail] = line[i];
    }
    while(head < tail && (cross ((Q[tail]&Q[tail-1]) - Q[head].s, Q[head].e-Q[head].s)) > eps)
        tail--;
    while(head < tail && (cross ((Q[head]&Q[head-1]) -Q[tail].s, Q[tail].e-Q[tail].e)) > eps) 
        head++;
    if(tail <= head + 1)
        return; 
    for(int i = head; i < tail; i++)
        res[resn++] = Q[i]&Q[i+1]; 
    if(head < tail - 1)
        res[resn++] = Q[head]&Q[tail];
}

int n, m;
Point p[maxn], ans[maxn];
Line hp[maxn];

bool ok (int x) {
    for (int i = 0; i < n; i++) {
        hp[i] = Line (p[i], p[(i+x+1)%n]);
    }
    HPI (hp, n, ans, m);
    if (fabs (polygon_area (ans, m)) >= eps)
        return 1;
    return 0;
}

int main () {
    int T;
    scanf ("%d", &T);
    while (T--) {
        scanf ("%d", &n);
        for (int i = n-1; i >= 0; i--) p[i].input ();
        if (n == 3) {
            printf ("1\n");
            continue;
        }
        int L = 1, R = n-2;
        while (R-L > 1) {
            int mid = (L+R)/2;
            if (ok (mid))
                L = mid;
            else 
                R = mid;
        }
        printf ("%d\n", ok (L) ? R : L);
    }
    return 0;
}