BZOJ 1007 HNOI 2008 水平可見直線 計算幾何+棧

題目大意：給出一些笛卡爾系中的一些直線，問從(0,+∞)向下看時能看到哪些直線。

思路：半平面交可做，可是顯然用不上。

類似於求凸包的思想，維護一個棧。

先將全部直線依照k值排序。然後挨個壓進去，遇到有前一個交點被擋住的話就先彈棧。

比較鬧心的是去重。我的方法是壓棧之前先去重，然後在處理。

CODE：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 50010
#define EPS 1e-5
using namespace std;

inline bool dcmp(double a,double b);

struct Point{
	double x,y;
	
	Point(double _ = 0,double __ = 0):x(_),y(__) {}
 	Point operator +(const Point &a)const {
		return Point(x + a.x,y + a.y);
	}
  	Point operator -(const Point &a)const {
		return Point(x - a.x,y - a.y);
	}
};
struct Line{
	double k,b;
	int _id;
	
	bool operator <(const Line &a)const {
		if(dcmp(k,a.k))	return b < a.b;
		return k < a.k;
	}
}line[MAX],_line[MAX];

int lines;
int top;
Line *stack[MAX]; 
Point intersection[MAX];

bool ans[MAX];

inline Point GetIntersection(const Line &l1,const Line &l2);
inline bool Under(const Line &l,const Point &p);

int main()
{
	cin >> lines;
	for(int i = 1;i <= lines; ++i) {
		scanf("%lf%lf",&line[i].k,&line[i].b);
		line[i]._id = i;
	}
	sort(line + 1,line + lines + 1);
	int cnt = 0;
	for(int i = 1;i <= lines; ++i) {
		if(dcmp(line[i].k,line[i + 1].k))	continue;
		_line[++cnt] = line[i];
	}
	stack[++top] = &_line[1];
	stack[++top] = &_line[2];
	intersection[top - 1] = GetIntersection(_line[1],_line[2]);
	for(int i = 3;i <= cnt; ++i) {
		if(i != cnt && _line[i].k == _line[i + 1].k)	continue;
		while(top > 1 && Under(_line[i],intersection[top - 1]))	--top;
		stack[++top] = &_line[i];
		intersection[top - 1] = GetIntersection(*stack[top],*stack[top - 1]);
	}
	for(int i = 1;i <= top; ++i)
		ans[stack[i]->_id] = true;
	for(int i = 1;i <= lines; ++i)
		if(ans[i])	printf("%d ",i);
	return 0;
}

inline bool dcmp(double a,double b)
{
	return fabs(a - b) < EPS;
}

inline Point GetIntersection(const Line &l1,const Line &l2)
{
	Point re;
	re.x = (l2.b - l1.b) / (l1.k - l2.k);
	re.y = re.x * l1.k + l1.b;
	return re;
}

inline bool Under(const Line &l,const Point &p)
{
	if(l.k * p.x + l.b > p.y || dcmp(l.k * p.x + l.b,p.y))	return true;
	return false;
}