BZOJ2732: [HNOI2012]射箭

列出不等式，二分，半平面交判定。注意可行域可能为点或线段或无界区域。坐标范围略大，算出来a的范围约为-1e9~-1e-18（a<0），理论上要用long double，不过好像double也能过。

#include<algorithm>
#include<cstdio>
#include<cmath>
using std::sort;
typedef long double flo;
const flo eps=1e-24;
const int N=2e5+5;
int m;
struct vec{flo x,y;};
flo det(vec a,vec b){return a.x*b.y-a.y*b.x;}
vec operator+(vec a,vec b){return(vec){a.x+b.x,a.y+b.y};}
vec operator-(vec a,vec b){return(vec){a.x-b.x,a.y-b.y};}
vec operator*(flo a,vec b){return(vec){a*b.x,a*b.y};}
struct line{
	int i;
	vec p,v;
	flo a;
	void cal(){a=atan2(v.y,v.x);}
}c[N],q[N];
flo cal(vec a,line b){return det(a-b.p,b.v);}
bool operator<(line a,line b){
	return a.a<b.a||a.a==b.a&&cal(a.p,b)<0;
}
vec over(line a,line b){
	return a.p+det(a.p-b.p,b.v)/det(b.v,a.v)*a.v;
}
bool jud(int s){
	int a=0,b=-1;
	for(int i=0;i<m;++i)
		if(c[i].i<=s)
			if(a>b||fabs(c[i].a-q[b].a)>eps){
				while(a<b&&cal(over(q[b],q[b-1]),c[i])>0)--b;
				while(a<b&&cal(over(q[a],q[a+1]),c[i])>0)++a;
				q[++b]=c[i];
			}
	while(a<b&&cal(over(q[b],q[b-1]),q[a])>0)--b;
	return b-a>1;
}
int main(){
	struct{
		operator int(){
			int x=0,c=getchar();
			while(c<48)c=getchar();
			while(c>47)
				x=x*10+c-48,c=getchar();
			return x;
		}
	}it;
	c[m++]=(line){0,-1e9,0,0,-1};
	c[m++]=(line){0,-1e-18,0,0,1};
	int n=it;
	for(int i=1;i<=n;++i){
		flo x=it,y1=it,y2=it;
		c[m++]=(line){i,0,y1/x,1,-x};
		c[m++]=(line){i,0,y2/x,-1,x};
	}
	for(int i=0;i<m;++i)
		c[i].cal();
	sort(c,c+m);
	int l=1,r=n;
	while(l!=r){
		int j=l+r+1>>1;
		jud(j)?l=j:r=j-1;
	}
	printf("%d
",l);
}