【poj 2653】 Pick-up sticks 线段相交

最新推荐文章于 2020-06-24 20:37:33 发布

Sun.R@in

最新推荐文章于 2020-06-24 20:37:33 发布

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文章标签： numbers input struct each random output

本文链接：https://blog.youkuaiyun.com/zfz1015/article/details/7889878

本文介绍了一个基于平面坐标的游戏——挑棍游戏。游戏的目标是找出所有顶部的棍子，即没有其他棍子覆盖的棍子。文章详细解释了游戏规则、输入输出格式及示例，并提供了一段C++代码实现来解决这个问题。

Pick-up sticks

Time Limit : 6000/3000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 5 Accepted Submission(s) : 5

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

Sample Output

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

Source

PKU

// 12-8-21.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include<cstdio>
#include<iostream>
using namespace std;
#define eps -1e-6
#define MAX 100005
struct Point
{
	double x,y;
};
struct Line
{
	Point a,b;
	int flag;
}s[MAX];
inline double cross(Point &o,Point &a, Point &b)  //oa*ob  (a.x-o.x,a.y-o.y)*(b.x-o.x,b.y-o.y)  *表示叉积
{  
    return   (a.x-o.x)*(b.y-o.y)-(a.y-o.y)*(b.x-o.x);
} 
inline bool quick(Line &p,Line &q)
{//一个快速排斥，一个跨立，推出线段相交
	if(min(p.a.x,p.b.x)<=max(q.a.x,q.b.x)&&
		min(q.a.x,q.b.x)<=max(p.a.x,p.b.x)&&
		min(p.a.y,p.b.y)<=max(q.a.y,q.b.y)&&
		min(q.a.y,q.b.y)<=max(p.a.y,p.b.y)&&
		 cross(p.a,q.a,q.b)*cross(p.b,q.a,q.b)<eps&&//验证p的两端点是否在q两端
		 cross(q.a,p.a,p.b)*cross(q.b,p.a,p.b)<eps)//验证q的两端点是否在p两端
		return true;
	return false;
}

int main()
{
	int n,i,j;
	while(scanf("%d",&n),n)
	{
		for(i=1;i<=n;i++)
		{
			scanf("%lf%lf%lf%lf",&s[i].a.x,&s[i].a.y,&s[i].b.x,&s[i].b.y);
			s[i].flag=1;
		}
		printf("Top sticks:");
		for(i=1;i<n;i++)
		{
			for(j=i;j<=n;j++)
				if(quick(s[i],s[j])) break;
			if(j==n+1) printf(" %d,", i);
		}	
		printf(" %d.\n",n);
	}
	return 0;
}