HDU1147 Pick-up sticks （线段相交）

题目描述

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.

分析

找所谓最上面的棍子，实际上是找线段相交。
求当前线段是否与之后的线段相交问题。如果没有，就认为在最上面。
线段相交解释

C++ 代码

#include<bits/stdc++.h>
using namespace std;
const int N = 1e5+10;
const int eps=1e-8;
struct node{
	double x,y;
}point1[N],point2[N];
int n;
bool st[N];
bool jdg(node &a,node &b,node &c,node &d)
{
    //快速排斥试验
	if(!(max(a.x,b.x)>=min(c.x,d.x) &&
		 max(c.x,d.x)>=min(a.x,b.x) &&
		 max(a.y,b.y)>=min(c.y,d.y) &&
		 max(c.y,d.y)>=min(a.y,b.y) ))
		return 0;
	
	//跨立实验
	double u,v,w,z;     
    u=(c.x-a.x)*(b.y-a.y)-(b.x-a.x)*(c.y-a.y);  
    v=(d.x-a.x)*(b.y-a.y)-(b.x-a.x)*(d.y-a.y);  
    w=(a.x-c.x)*(d.y-c.y)-(d.x-c.x)*(a.y-c.y);  
    z=(b.x-c.x)*(d.y-c.y)-(d.x-c.x)*(b.y-c.y);  
    return (u*v<=eps && w*z<=eps);
}
int main()
{
	while(scanf("%d",&n))
	{
		if(n==0) break;
		memset(st,0,sizeof st);
		for(int i=1;i<=n;i++)
		{
			scanf("%lf%lf%lf%lf",&point1[i].x,&point1[i].y,&point2[i].x,&point2[i].y);
		}
		for(int i=1;i<=n;i++)
		{
			for(int j=i+1;j<=n;j++)
			{
				if(jdg(point1[i],point2[i],point1[j],point2[j]))    //如果当前的线段与之后的线段相交
				{                                                   //那它肯定不在最上面，退出循环
					st[i]=1;
					break;
				}
			}
		}
		printf("Top sticks: ");
		bool f=false;
		for(int i=1;i<=n;i++)
		{
			if(!st[i])
			{
				if(f) printf(", "); 
				printf("%d",i);
				f=true;
			}
		}
		printf(".\n");
	}
    return 0;
}