题目大意:给出一些笛卡尔系中的一些直线,问从(0,+∞)向下看时能看到哪些直线。
思路:半平面交可做,但是显然用不上。类似于求凸包的思想,维护一个栈。先将所有直线按照k值排序,然后挨个压进去,遇到有前一个交点被挡住的话就先弹栈。
比较闹心的是去重。我的方法是压栈之前先去重,然后在处理。
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; }
BZOJ 1007 HNOI 2008 水平可见直线 计算几何+栈
原文地址:http://blog.csdn.net/jiangyuze831/article/details/40188269