bzoj 1007: [HNOI2008]水平可见直线 半平面交

题目大意:

http://www.lydsy.com/JudgeOnline/problem.php?id=1007;

题解

其实就是求每条直线的上半部分的交
所以做裸半平面交即可

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
inline void read(int &x){
	x=0;char ch;bool flag = false;
	while(ch=getchar(),ch<'!');if(ch == '-') ch=getchar(),flag = true;
	while(x=10*x+ch-'0',ch=getchar(),ch>'!');if(flag) x=-x;
}
inline int cat_max(const int &a,const int &b){return a>b ? a:b;}
inline int cat_min(const int &a,const int &b){return a<b ? a:b;}
const int maxn = 50010;
const double eps = 1e-9;
inline int dcmp(const double &x){
	if(x < eps && x > -eps) return 0;
	return x > 0 ? 1 : -1;
}
struct Point{
	double x,y;
	Point(){}
	Point(const double &a,const double &b){x=a;y=b;}
	void print(){
		printf("%lf %lf
",x,y);
	}
};
struct line{
	double k,b;
	int id;
};
inline bool cmp(const line &a,const line &b){
	return dcmp(a.k-b.k) == 0 ? a.b > b.b : a.k < b.k;
}
inline bool kmp(const line &a,const line &b){
	return a.id < b.id;
}
inline Point Interion(const line &x,const line &y){
	double t = (y.b - x.b)/(x.k - y.k);
	return Point(t,x.k*t+x.b);
}
line lines[maxn],sta[maxn];
int top = 0;
int main(){
	int n;read(n);
	for(int i=1;i<=n;++i){
		scanf("%lf%lf",&lines[i].k,&lines[i].b);
		lines[i].id = i;
	}sort(lines+1,lines+n+1,cmp);
	for(int i=1;i<=n;++i){
		if(dcmp(lines[i].k-sta[top].k) == 0) continue;
		while(top >= 2){
			Point x = Interion(lines[i],sta[top]);
			Point y = Interion(sta[top],sta[top-1]);
			if(dcmp(x.x-y.x) <= 0) --top;
			else break;
		}sta[++top] = lines[i];
	}sort(sta+1,sta+top+1,kmp);
	for(int i=1;i<=top;++i) printf("%d ",sta[i].id);
	getchar();getchar();
	return 0;
}