POJ3304（计算几何基础-判断线段与直线相交）

最新推荐文章于 2022-06-18 12:22:14 发布

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最新推荐文章于 2022-06-18 12:22:14 发布 · 461 阅读

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该博客探讨了如何确定是否存在一条直线，使得给定的n条线段在其投影上至少有一个共同点。通过计算几何的方法，将问题转化为判断每条线段是否与特定直线相交。给出了样例输入和输出，并提供了解题思路，即枚举线段端点以检查相交情况。

Segments
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 14774 Accepted: 4690
Description

Given n segments in the two dimensional space, write a program, which determines if there exists a line such that after projecting these segments on it, all projected segments have at least one point in common.

Input

Input begins with a number T showing the number of test cases and then, T test cases follow. Each test case begins with a line containing a positive integer n ≤ 100 showing the number of segments. After that, n lines containing four real numbers x1 y1 x2 y2 follow, in which (x1, y1) and (x2, y2) are the coordinates of the two endpoints for one of the segments.

Output

For each test case, your program must output “Yes!”, if a line with desired property exists and must output “No!” otherwise. You must assume that two floating point numbers a and b are equal if |a - b| < 10-8.

Sample Input

3
2
1.0 2.0 3.0 4.0
4.0 5.0 6.0 7.0
3
0.0 0.0 0.0 1.0
0.0 1.0 0.0 2.0
1.0 1.0 2.0 1.0
3
0.0 0.0 0.0 1.0
0.0 2.0 0.0 3.0
1.0 1.0 2.0 1.0
Sample Output

Yes!
Yes!
No!
Source

Amirkabir University of Technology Local Contest 2006

题意：给你n条线段，问你存不存在一条直线，使得这n条线段在这条直线上的投影至少有一个公共点。

解题思路：问题可以转换为存不存在一条直线与n条线段都相交，那么我们要求的线段就与这条线段垂直，那么我们要找一条直线与这n条线段都相交，只需要枚举所有两条线段的端点（也要枚举一条线段自身的两个端点），然后判断这两个端点组成的直线与n条线段是否相交。

#include<stdio.h>
#include<iostream>
#include<cmath>
#include<algorithm>

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