Pick-up sticks

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Problem Description

Stan has n sticks of various length. He throws them one at a time on the floor in a random way. After finishing throwing, Stan tries to find the top sticks, that is these sticks such that there is no stick on top of them. Stan has
noticed that the last thrown stick is always on top but he wants to know all the sticks that are on top. Stan sticks are very, very thin such that their thickness can be neglected.

Input

Input consists of a number of cases. The data for each case start with 1 ≤ n ≤ 100000, the number of sticks for this case. The following n lines contain four numbers each, these numbers are the planar coordinates of the endpoints
of one stick. The sticks are listed in the order in which Stan has thrown them. You may assume that there are no more than 1000 top sticks. The input is ended by the case with n=0. This case should not be processed.

Output

For each input case, print one line of output listing the top sticks in the format given in the sample. The top sticks should be listed in order in which they were thrown.

The picture to the right below illustrates the first case from input.

Sample Input

5
1 1 4 2
2 3 3 1
1 -2.0 8 4
1 4 8 2
3 3 6 -2.0
3
0 0 1 1
1 0 2 1
2 0 3 1
0

Sample Output

Top sticks: 2, 4, 5.
Top sticks: 1, 2, 3.

代码：

#include <stdio.h>
#define MAX 100100

struct Point{
	double x ,y ;
	Point operator-(Point p)
	{
		Point ans ;
		ans.x = x-p.x ;
		ans.y = y-p.y ;
		return ans ;
	}
};

struct Segment{
	Point s , e ;
	bool cover ;
}seg[MAX];

double min(double x , double y)
{
	return x>y?y:x ;
}
double max(double x , double y)
{
	return x>y?x:y ;
}
bool quickExclude(const Segment &a , const Segment &b)
{
	double minRX = min(a.s.x,a.e.x) , minRY = min(a.s.y,a.e.y) ;
	double maxRX = max(a.s.x,a.e.x) , maxRY = max(a.s.y,a.e.y) ;
	double minTX = min(b.s.x,b.e.x) , minTY = min(b.s.y,b.e.y) ;
	double maxTX = max(b.s.x,b.e.x) , maxTY = max(b.s.y,b.e.y) ;
	if(max(minTX,minRX)>min(maxTX,maxRX) && max(minTY,minRY)>min(maxTY,maxRY))
	{
		return false ;
	}
	return true ;
}

double f(Point p1 , Point p2)
{
	return p1.x*p2.y-p1.y*p2.x ;
}

bool ifIntersect(Segment &a , Segment &b)
{
	if(quickExclude(a,b))
	{
		if(f(a.s-b.s,b.e-b.s)*f(b.e-b.s,a.e-b.s)>=0 &&
			f(b.s-a.s,a.e-a.s)*f(a.e-a.s,b.e-a.s)>=0 )
			{
				return true ;
			}
	}
	return false ;
}

int main()
{
	int n ;
	while(~scanf("%d",&n) && n)
	{
		for(int i = 0 ; i < n ; ++i)
		{
			scanf("%lf%lf%lf%lf",&seg[i].s.x,&seg[i].s.y,&seg[i].e.x,&seg[i].e.y) ;
			seg[i].cover = false ;
		}
		for(int i = 0 ; i < n ; ++i) 		//注意只能这样写！！下面有超时的代码例子
		{
			for(int j = i+1 ; j < n ; ++j)
			{
				if(ifIntersect(seg[i],seg[j]))
				{
					seg[i].cover = true ;
					break ;
				}
			}
		}
		/*  超时的代码！！！
		 for(int i = 0 ; i < n ; ++i)
		{
			for(int j = i+1 ; j < n ; ++j)
			{
				if(!seg[i].cover)
				{
					if(ifIntersect(seg[i],seg[j]))
					{
						seg[i].cover = true ;
						break ;
					}
				}
			}
		}
		*/
		int i = 0;
		for(i = 0 ; i < n ; ++i)
		{
			if(!seg[i].cover)
			{
				printf("Top sticks: %d",i+1) ;
				break ;
			}
		}
		for(i=i+1; i < n ; ++i)
		{
			if(!seg[i].cover)
			{
				printf(", %d",i+1) ;
			}
		}
		puts(".") ;
	}
	return 0 ;
}

与君共勉