POJ 2653 Pick-up sticks (判断线段相交)_poj 2653 pick-up sticks-优快云博客

本文链接：https://blog.youkuaiyun.com/helloiamclh/article/details/49643377

本文介绍了一个解决Pick-upsticks问题的算法实现过程，包括输入格式、问题描述、解题思路及AC代码，详细解释了如何通过倒序遍历棍子并检查相交情况来找出最终的顶层棍子。

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Pick-up sticks

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 11329		Accepted: 4247

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

Sample Output

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




题意：n个棍子，一个一个扔，如果扔在了别的棍子上，也就是说和别的棍子相交了，那么和它相交的棍子就会消失，

求最后剩下的棍子的编号

思路：一个一个遍历，但是倒着检查，如果它的上面有棍子，那么这根棍子就一定会消失，然后继续下一个棍子

ac代码：
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#define MAXN 101000
#define MOD 1000000007
#define LL long long
#define INF 0xfffffff
#define fab(a)(a)>0?(a):(-a)
using namespace std;
struct s
{
	double x1,x2,y1,y2;
}a[MAXN];
int v[MAXN];
double fun(s aa,double xx,double yy)
{
	return (aa.x2-aa.x1)*(yy-aa.y1)-(xx-aa.x1)*(aa.y2-aa.y1);
}
int check(s A,s B)//检查相交
{
	if(max(A.x1,A.x2)<min(B.x1,B.x2))
	return 0;
	if(max(B.x1,B.x2)<min(A.x1,A.x2))
	return 0;
	if(max(A.y1,A.y2)<min(B.y1,B.y2))
	return 0;
	if(max(B.y1,B.y2)<min(A.y1,A.y2))
	return 0;
	if(fun(B,A.x1,A.y1)*fun(B,A.x2,A.y2)>=1e-10)
	return 0;
	if(fun(A,B.x1,B.y1)*fun(A,B.x2,B.y2)>=1e-10)
	return 0;
	return 1;
}
int main()
{
    int n,i,j;
	while(scanf("%d",&n)!=EOF,n)
	{
		memset(v,0,sizeof(v));
		for(i=1;i<=n;i++)
		{
			scanf("%lf%lf%lf%lf",&a[i].x1,&a[i].y1,&a[i].x2,&a[i].y2);
		}
		printf("Top sticks: ");
		for(i=1;i<n;i++)
		{
			for(j=i+1;j<=n;j++)//倒着检查
			{
				if(check(a[i],a[j]))
				{
					break;
				}
			}
			if(j==n+1)
			printf("%d, ",i);
		}
		printf("%d.\n",n);
	} 
    return 0;
}