SGU 253. Theodore Roosevelt

Link To The Problem

Solution : 三角剖分、二分、极角排序

Code:

//SGU 253 Theodore Roosevelt
//Solution : 三角剖分、二分

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#include<queue>

using namespace std;
#define FOR(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)
#define DOR(i,a,b) for(int (i)=(a);(i)>=(b);(i)--)

#define oo 1e6
#define eps 1e-8
#define nMax 101000

//{ 
#define pb push_back
#define dbg(x) cerr << __LINE__ << ": " << #x << " = " << x << endl

#define F first
#define S second

#define bug puts("OOOOh.....");
#define zero(x) (((x)>0?(x):-(x))<eps)

#define LL long long
#define DB double 

#define sf scanf
#define pf printf
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)

double const pi = acos(-1.0);
double const inf = 1e9;
double inline sqr(double x) { return x*x; }

int dcmp(double x){
    if(fabs(x)<=eps) return 0;
    return x>0?1:-1;
}
//}

// Describe of the 2_K Geomtry
// First Part : Point and Line
// Second Part Cicle
// Third Part Polygan

// First Part:
// ****************************** Point and Line *******************************\\
//  {  

class point {
public:
    double x,y;
    point (double x=0,double y=0):x(x),y(y) {}
    void make(double _x,double _y) {x=_x;y=_y;}
    void read() { scanf("%lf%lf",&x,&y); }
    void out() { printf("%.3lf %.3lf\n",x,y);}
    double len() { return sqrt(x*x+y*y); }

    point friend operator - (point const& u,point const& v) { return point(u.x-v.x,u.y-v.y); }
    point friend operator + (point const& u,point const& v) { return point(u.x+v.x,u.y+v.y); }
    double friend operator * (point const& u,point const& v){ return u.x*v.y-u.y*v.x; }
    double friend operator ^ (point const& u,point const& v) { return u.x*v.x+u.y*v.y; }
    point friend operator * (point const& u,double const& k) { return point(u.x*k,u.y*k); }
	point friend operator / (point const& u,double const& k) { return point(u.x/k,u.y/k); }
	
	friend bool operator < (point const& u,point const& v){
		if(dcmp(v.x-u.x)==0) return dcmp(u.y-v.y)<0;
		return dcmp(u.x-v.x)<0;
	}
	friend bool operator != (point const& u,point const& v){
		return dcmp(u.x-v.x) || dcmp(u.y-v.y);
	}
	friend bool operator == (point const& u,point const& v){
		return dcmp(u.x-v.x)==0 && dcmp(u.y-v.y)==0;
	}

	point rotate(double s) {
		return point(x*cos(s) + y*sin(s),\
					-x*sin(s) + y*cos(s));
	}
};
typedef point Vector;
typedef class line{
public:
    point a,b;
    line() {}
    line (point a,point b):a(a),b(b){}
    void make(point u,point v) {a=u;b=v;}
    void read() { a.read(),b.read(); }
}segment;

double det(point u,point v) {
	return u.x*v.y - u.y*v.x;
}
double dot(point u,point v) {
	return u.x*v.x + u.y*v.y;
}

// Weather P is On the Segment (uv) 
int dot_on_seg(point p,point u,point v){
	return dcmp(det(p-u,v-p))==0 && dcmp(dot(p-u,p-v)) <= 0; // '>=' means P is u or v
}
// The distance from point p to line l
double PToLine(point p,line l) {
	return fabs((p-l.a)*(l.a-l.b))/(l.a-l.b).len();
}
// The ProJect Of Point(p) To Line(l)
point PointProjectLine(point p,line l) {
	double t = dot(l.b-l.a,p-l.a)/dot(l.b-l.a,l.b-l.a);
	return l.a + (l.b-l.a)*t;
}
// Weather line u parallel line v
int parallel(line u,line v) {
	return dcmp(det(u.a-u.b,v.a-v.b))==0;
}
// The Intersection Point Of Line u and Line v
point intersection(line u,line v) {
	point ret = u.a;
	double t = det(u.a-v.a,v.a-v.b)/det(u.a-u.b,v.a-v.b);
	return ret + (u.b-u.a)*t;
}

//}
// ****************************** First Part end ********************************\\

// Second Part:
// ********************************* Circle *************************************\\
// {

struct Circle {
	point O;
	double r;
	Circle() {};
	Circle(point O,double r):O(O),r(r){};
};


//}
// ****************************** Second Part End *******************************\\

// Third Part :
// ********************************* Polygan *************************************\\
// {


int ConvexHull(vector<point>& p){
	int n=p.size();
	int m=0;
	vector<point> q;
	q.resize(2*n+5);
	rep(i,n) {
		while(m>1 && dcmp((q[m-1]-q[m-2])*(p[i]-q[m-2])) <= 0) m--;
		q[m++] = p[i];
	}
	int k = m;
	for(int i=n-2;i>=0;i--) {
		while(m>k && dcmp((q[m-1]-q[m-2])*(p[i]-q[m-2])) <= 0) m--;
		q[m++] = p[i];
	}
	q.resize(m) ;
	if(m>1) q.resize(m-1);
	// p = q;    // 是否修改原来的多边形
	return q.size();
}
// 三角形重心
point Center(point a,point b,point c){
	return (a+b+c)/3.0;
}
// Centroid of Polygan
point Centroid(vector<point> p){
	point O(0,0),ret(0,0);
	int n = p.size();
	p.pb(p[0]);
	double area = 0.0;
	rep(i,n) {
		ret = ret + Center(O,p[i],p[i+1])*dot(p[i]-O,p[i+1]-O);
		area += dot(p[i]-O,p[i+1]-O);
	}
	if(dcmp(area)==0) {
		pf("There maybe something wrong\n");
		return p[0];
	}
	return ret / area;
}


struct Polygan{
	vector<point> g;
	Polygan() {};
	Polygan(vector<point> g):g(g){};
	Polygan(point p[],int n) {g.clear();rep(i,n) g.pb(p[i]); };
	int convex() { return ConvexHull(g); }
	point center() { return Centroid(g);  } // 多边形的重心
};

//}
// ******************************* Third Part End ********************************\\


// Solution Part :

int dot_in_Tri(point p,point a,point b,point c){
	return dcmp((c-b)*(p-b))>=0;
}
double Angle(point p) {
	return atan2(p.y,p.x);
}

int n,m,k;
point p[nMax];
vector<point> a;
point s[nMax];
vector<double> ang;
point O;

int find(point q) {
	int l=0,r=a.size()-1;
	int ans=-1,mid;
	if(q < O) return 0;
	if(q == O) return 1;
	while(l<=r) {
		mid = (l+r)>>1;
		if(Angle((q-O))>=ang[mid]){
			ans = mid;
			l = mid+1;
		}else r = mid-1;
	}
//	dbg(ans);
	if(ans == -1) return 0;
	if(ans==a.size()-1){
		return dot_on_seg(q,O,a[a.size()-1]);
	}
	return dot_in_Tri(q,O,a[ans],a[ans+1]);
}
void init(int cur) {
	a.clear() ;
	ang.clear();
	for(int i=cur+1;i<n;i++) {
		a.pb(p[i]);
		ang.pb(Angle(p[i]-O));
	}
	for(int i=0;i<cur;i++) {
		a.pb(p[i]);
		ang.pb(Angle(p[i]-O));
	}
//	rep(i,ang.size()) pf("%.3lf ",ang[i]);pf("\n");
}
int main() {
#ifndef ONLINE_JUDGE
	freopen("in.txt","r",stdin);
#endif
	while(~sf("%d%d%d",&n,&m,&k)){
		O = point(inf,inf);
		int cur = -1;
		for(int i=0;i<n;i++) {
			p[i].read() ;
			if(p[i] < O) {
				O = p[i];
				cur = i;
			}
		}
		init(cur);
		rep(i,m) s[i].read();
		int cnt = 0;
		for(int i=0;i<m;i++) {
			cnt += find(s[i]);
			//dbg(i);
			//bug
		}
		//cout << cnt << endl;
		pf("%s\n",cnt>=k?"YES":"NO");
	}
	return 0;
}