POJ2653 Pick-up sticks 判断线段相交

POJ2653

判断线段相交的方法 先判断直线是否相交 再判断点是否在线段上 复杂度是常数的

题目保证最后答案小于1000

故从后往前尝试用后面的线段 "压"前面的线段 排除不可能的答案 就可以轻松AC了。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;

const double eps=1e-9;

int cmp(double x)
{
 if(fabs(x)<eps)return 0;
 if(x>0)return 1;
 	else return -1;
}

const double pi=acos(-1.0);

inline double sqr(double x)
{
 return x*x;
}

struct point
{
 double x,y;
 point (){}
 point (double a,double b):x(a),y(b){}
 void input()
 	{
 	 scanf("%lf%lf",&x,&y);
	}
 friend point operator +(const point &a,const point &b)
 	{
 	 return point(a.x+b.x,a.y+b.y);
	}	
 friend point operator -(const point &a,const point &b)
 	{
 	 return point(a.x-b.x,a.y-b.y);
	}
 friend bool operator ==(const point &a,const point &b)
 	{
 	 return cmp(a.x-b.x)==0&&cmp(a.y-b.y)==0;
	}
 friend point operator *(const point &a,const double &b)
 	{
 	 return point(a.x*b,a.y*b);
	}
 friend point operator*(const double &a,const point &b)
 	{
 	 return point(a*b.x,a*b.y);
	}
 friend point operator /(const point &a,const double &b)
 	{
 	 return point(a.x/b,a.y/b);
	}
 double norm()
 	{
 	 return sqrt(sqr(x)+sqr(y));
	}
};

double det(const point &a,const point &b)
{
 return a.x*b.y-a.y*b.x;
}

double dot(const point &a,const point &b)
{
 return a.x*b.x+a.y*b.y; 
}

double dist(const point &a,const point &b)
{
 return (a-b).norm();
}

point rotate_point(const point &p,double A)
{
 double tx=p.x,ty=p.y;
 return point(tx*cos(A)-ty*sin(A),tx*sin(A)+ty*cos(A));
}

struct line
{
 point a,b;
 line(){};
 line(point x,point y):a(x),b(y)
 {
 	
 }
};


bool parallel(line a,line b)
{
 return !cmp(det(a.a-a.b,b.a-b.b));
}

bool line_joined(line a,line b,point &res)
{
 if(parallel(a,b))return false;
 double s1=det(a.a-b.a,b.b-b.a);
 double s2=det(a.b-b.a,b.b-b.a);
 res=(s1*a.b-s2*a.a)/(s1-s2);
 return true;
}

bool pointonSegment(point p,point s,point t)
{
 return cmp(det(p-s,t-s))==0&&cmp(dot(p-s,p-t))<=0;
}

const int maxn=100000+1;
line li[maxn];
deque<int>q;
deque<int>qq;
int main()
{freopen("t.txt","r",stdin);
 //freopen("1.txt","w",stdout);
 int n;
 while(scanf("%d",&n))
 	{
 	 while(!q.empty())q.pop_front();
 	 while(!qq.empty())qq.pop_front();
 	 if(n==0)return 0;
 	 for(int i=1;i<=n;i++)
 	 	{
 	 	 point a,b;
 	 	 a.input();b.input();
 	 	 li[i]=line(a,b);
 	 	 
		 q.push_back(i);
		}
	 for(int i=n;i>1;i--)
	 	{
	 	  while(!q.empty())
 	 	 	{
 	 	 	 
 	 	 	 int nv=q.front();
 	 	 	 if(nv>=i)break;
 	 	 	 line nl=li[nv];q.pop_front();
 	 	 	 point jo;
 	 	 	 if(!line_joined(nl,li[i],jo)){qq.push_back(nv);continue;}
 	 	 	 if((pointonSegment(jo,nl.a,nl.b))&&(pointonSegment(jo,li[i].a,li[i].b)))continue;
			 qq.push_back(nv);
			}
		 while(!qq.empty())
		 	{
			  int nv=qq.back();qq.pop_back();
			  q.push_front(nv);
			}
		}
	 printf("Top sticks: ");
	 while(!q.empty())
	 	{
	 	 int now=q.front();q.pop_front();
	 	 if(q.empty())printf("%d.",now);
	 	 	else printf("%d, ",now);
		}
	 printf("
");
	}
 return 0;
}