【POJ1912】【Ceoi2002】—A highway and the seven dwarfs（凸包）

传送门

题意：给出平面上的 $n$ 个点，对于 $m$ 条直线，依次判断这 $n$ 个点是否在每条直线的同一侧。($n,m≤10^5$)

考虑对于每一条直线，显然我们要找到凸包上距离这条直线最远的2个点，判一下方向

考虑到类似于旋转卡壳对踵的的思想
记录一下每条边的斜率，然后二分找到的就是最远的点了

#include<bits/stdc++.h>
using namespace std;
inline int read(){
	char ch=getchar();
	int res=0,f=1;
	while(!isdigit(ch)){if(ch=='-')f=-f;ch=getchar();}
	while(isdigit(ch))res=res*10+(ch^48),ch=getchar();
	return res*f;
}
const int N=100005;
const double pi=acos(-1);
const double eps=1e-8;
struct point{
	double x,y;
	point(double a=0,double b=0){
		x=a,y=b;
	}
	friend inline point operator +(const point &a,const point &b){
		return point(a.x+b.x,a.y+b.y);
	}
	friend inline point operator -(const point &a,const point &b){
		return point(a.x-b.x,a.y-b.y);
	}
	friend inline double operator *(const point &a,const point &b){
		return (a.x*b.y-a.y*b.x);
	}
	inline int calc(){
		return x*x+y*y;
	}
}p[N],q[N],a,b;
int n,top;
double af[N];
inline bool comp(const point &a,const point &b){
	double res=(a-p[1])*(b-p[1]);
	return (res==0)?((a-p[1]).calc()<(b-p[1]).calc()):(res>0);
}
inline void graham(){
	int idx=1;
	for(int i=2;i<=n;i++){
		if(p[idx].y<p[i].y||(p[idx].y==p[i].y&&p[i].x<p[idx].x))
		idx=i;
	}
	if(idx!=1)swap(p[idx],p[1]);
	sort(p+2,p+n+1,comp);
	q[++top]=p[1];
	for(int i=2;i<=n;++i){
		while(top>=3&&((p[i]-q[top-1])*(q[top]-q[top-1])>=0))
			top--;
		q[++top]=p[i];
	}
	q[top+1]=q[1];
	for(int i=1;i<=top;i++)
		af[i]=atan2(q[i+1].y-q[i].y,q[i+1].x-q[i].x);
}
int main(){
	n=read();
	for(int i=1;i<=n;i++){
		scanf("%lf%lf",&p[i].x,&p[i].y);
	}
	if(n>1)graham();
	while(scanf("%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y)!=EOF){
		if(n==1){
			cout<<"GOOD"<<'\n';continue;
		}
		double l1=atan2(b.y-a.y,b.x-a.x),l2=atan2(a.y-b.y,a.x-b.x);
		int pos1=lower_bound(af+1,af+top+1,l1)-af;
		int pos2=lower_bound(af+1,af+top+1,l2)-af;
		point c=q[pos1],d=q[pos2];
		if(((b-a)*(c-a))*((b-a)*(d-a))<=0&&(((d-c)*(a-c))*((d-c)*(b-c))))
			cout<<"BAD"<<'\n';
		else cout<<"GOOD"<<'\n';
	}
}