●poj 1474 Video Surveillance

Posted 2020-10-22 *ZJ

题链：

http://poj.org/problem?id=1474

题解：

计算几何，半平面交

半平面交裸题，快要恶心死我啦。。。

（了无数次之后，一怒之下把onleft改为onright，然后还加了一个去重，总算是过了。。。）

代码：

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define MAXN 1550
using namespace std;
const double eps=1e-8;
int sign(double x){
	if(fabs(x)<=eps) return 0;
	return x<0?-1:1;
}
struct Point{
	double x,y;
	Point(double _x=0,double _y=0):x(_x),y(_y){}
	void Read(){scanf("%lf%lf",&x,&y);}
}D[MAXN];
typedef Point Vector;
struct Line{
	Point s; Vector v; double ang;
	Line(){}
	Line(Point _s,Vector _v):s(_s),v(_v){ang=atan2(v.y,v.x);}
	friend bool operator < (Line A,Line B){return sign(A.ang-B.ang)<0;}
}L[MAXN];
bool operator == (Point A,Point B){return sign(A.x-B.x)==0&&sign(A.y-B.y)==0;}
Vector operator + (Vector A,Vector B){return Vector(A.x+B.x,A.y+B.y);}
Vector operator - (Point A,Point B){return Vector(A.x-B.x,A.y-B.y);}
Vector operator * (Vector A,double k){return Vector(A.x*k,A.y*k);}
double operator ^ (Vector A,Vector B){return A.x*B.y-A.y*B.x;}
double operator * (Vector A,Vector B){return A.x*B.x+A.y*B.y;}
bool OnRight(Point P,Line A){return sign(A.v^(P-A.s))<0;}
Point LLI(Line A,Line B){//Line_Line_Intersection
	Vector u=B.s-A.s;
	double t1=(u^B.v)/(A.v^B.v);
	return A.s+A.v*t1;
}
int HPI(int lnt){//Half_Plane_Intersection
	static Point QP[MAXN]; static Line QL[MAXN];
	sort(L+1,L+lnt+1);
	int l=1,r=1,dnt=0; QL[r]=L[1];
	for(int i=2;i<=lnt;i++){
		while(l<r&&OnRight(QP[r-1],L[i])) r--;
		while(l<r&&OnRight(QP[l],L[i])) l++;
		QL[++r]=L[i];
		if(sign(QL[r-1].v*QL[r].v)>0&&sign(QL[r-1].v^QL[r].v)==0){
			r--; if(!OnRight(L[i].s,QL[r])) QL[r]=L[i];
		}
		if(l<r) QP[r-1]=LLI(QL[r-1],QL[r]);
	}
	while(l<r&&OnRight(QP[r-1],QL[l])) r--;
	if(r-l<=1) return 0;
	QP[r]=LLI(QL[r],QL[l]);
	for(int i=l;i<=r;i++) D[++dnt]=QP[i];
	dnt=unique(D+1,D+dnt+1)-D-1;
	return dnt-(dnt>1&&D[dnt]==D[1]);
}
int main(){
	int n,Case=0; 
	while(~scanf("%d",&n)&&n){
		for(int i=1;i<=n;i++)
			D[i].Read(),L[i]=Line(D[i],D[i-1]-D[i]);
		L[1]=Line(D[1],D[n]-D[1]);
		n=HPI(n);
		printf("Floor #%d\n",++Case);
		//printf("%d\n",n);
		if(n>0) printf("Surveillance is possible.\n");
		else printf("Surveillance is impossible.\n");
		printf("\n");
	}
	return 0;
}