Intersecting Lines
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 15145   Accepted: 6640

Description

We all know that a pair of distinct points on a plane defines a line and that a pair of lines on a plane will intersect in one of three ways: 1) no intersection because they are parallel, 2) intersect in a line because they are on top of one another (i.e. they are the same line), 3) intersect in a point. In this problem you will use your algebraic knowledge to create a program that determines how and where two lines intersect. 
Your program will repeatedly read in four points that define two lines in the x-y plane and determine how and where the lines intersect. All numbers required by this problem will be reasonable, say between -1000 and 1000. 

Input

The first line contains an integer N between 1 and 10 describing how many pairs of lines are represented. The next N lines will each contain eight integers. These integers represent the coordinates of four points on the plane in the order x1y1x2y2x3y3x4y4. Thus each of these input lines represents two lines on the plane: the line through (x1,y1) and (x2,y2) and the line through (x3,y3) and (x4,y4). The point (x1,y1) is always distinct from (x2,y2). Likewise with (x3,y3) and (x4,y4).

Output

There should be N+2 lines of output. The first line of output should read INTERSECTING LINES OUTPUT. There will then be one line of output for each pair of planar lines represented by a line of input, describing how the lines intersect: none, line, or point. If the intersection is a point then your program should output the x and y coordinates of the point, correct to two decimal places. The final line of output should read "END OF OUTPUT". 

Sample Input

5
0 0 4 4 0 4 4 0
5 0 7 6 1 0 2 3
5 0 7 6 3 -6 4 -3
2 0 2 27 1 5 18 5
0 3 4 0 1 2 2 5

Sample Output

INTERSECTING LINES OUTPUT
POINT 2.00 2.00
NONE
LINE
POINT 2.00 5.00
POINT 1.07 2.20
END OF OUTPUT

Source


呵呵 这种裸题...
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
typedef long long ll;
const double eps=1e-;
inline int read(){
char c=getchar();int x=,f=;
while(c<''||c>''){if(c=='-')f=-; c=getchar();}
while(c>=''&&c<=''){x=x*+c-''; c=getchar();}
return x*f;
}
inline int sgn(double x){
if(abs(x)<eps) return ;
else return x<?-:;
}
struct Vector{
double x,y;
Vector(double a=,double b=):x(a),y(b){}
bool operator <(const Vector &a)const{
return x<a.x||(x==a.x&&y<a.y);
}
void print(){
printf("%lf %lf\n",x,y);
}
};
typedef Vector Point;
Vector operator +(Vector a,Vector b){return Vector(a.x+b.x,a.y+b.y);}
Vector operator -(Vector a,Vector b){return Vector(a.x-b.x,a.y-b.y);}
Vector operator *(Vector a,double b){return Vector(a.x*b,a.y*b);}
Vector operator /(Vector a,double b){return Vector(a.x/b,a.y/b);}
bool operator ==(Vector a,Vector b){return sgn(a.x-b.x)==&&sgn(a.y-b.y)==;} double Cross(Vector a,Vector b){
return a.x*b.y-a.y*b.x;
}
double Dot(Vector a,Vector b){
return a.x*b.x+a.y*b.y;
}
double DisPP(Point a,Point b){
Point t=a-b;
return sqrt(t.x*t.x+t.y*t.y);
}
struct Line{
Point s,t;
Line(){}
Line(Point p,Point v):s(p),t(v){}
};
bool isLSI(Line l1,Line l2){
Vector v=l1.t-l1.s,u=l2.s-l1.s,w=l2.t-l1.s;
return sgn(Cross(v,u))!=sgn(Cross(v,w));
}
bool isSSI(Line l1,Line l2){
return isLSI(l1,l2)&&isLSI(l2,l1);
}
Point LI(Line a,Line b){
Vector v=a.s-b.s,v1=a.t-a.s,v2=b.t-b.s;
double t=Cross(v2,v)/Cross(v1,v2);
return a.s+v1*t;
}
double x,y,x2,y2;
Line l1,l2;
void solve(){
if(sgn(Cross(l1.t-l1.s,l2.t-l2.s))==){
Vector v=l1.t-l1.s,u=l2.s-l1.s,w=l2.t-l1.s;
if(sgn(Cross(v,u))==&&sgn(Cross(v,w))==) puts("LINE");
else puts("NONE");
}else{
Point p=LI(l1,l2);
printf("POINT %.2f %.2f\n",p.x,p.y);
}
}
int main(int argc, const char * argv[]) {
int T=read();
puts("INTERSECTING LINES OUTPUT");
while(T--){
scanf("%lf%lf%lf%lf",&x,&y,&x2,&y2);
l1=Line(Point(x,y),Point(x2,y2));
scanf("%lf%lf%lf%lf",&x,&y,&x2,&y2);
l2=Line(Point(x,y),Point(x2,y2));
solve();
}
puts("END OF OUTPUT");
return ;
}
 

POJ1269 Intersecting Lines[线段相交 交点]的更多相关文章

  1. [poj1269]Intersecting Lines

    题目大意:求两条直线的交点坐标. 解题关键:叉积的运用. 证明: 直线的一般方程为$F(x) = ax + by + c = 0$.既然我们已经知道直线的两个点,假设为$(x_0,y_0), (x_1 ...

  2. POJ1269:Intersecting Lines(判断两条直线的关系)

    题目:POJ1269 题意:给你两条直线的坐标,判断两条直线是否共线.平行.相交,若相交,求出交点. 思路:直线相交判断.如果相交求交点. 首先先判断是否共线,之后判断是否平行,如果都不是就直接求交点 ...

  3. poj 1269 Intersecting Lines(直线相交)

    Intersecting Lines Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 8637   Accepted: 391 ...

  4. poj1269 intersecting lines【计算几何】

    We all know that a pair of distinct points on a plane defines a line and that a pair of lines on a p ...

  5. POJ1269 Intersecting Lines 2017-04-16 19:43 50人阅读 评论(0) 收藏

    Intersecting Lines Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 15478   Accepted: 67 ...

  6. Pipe - POJ 1039(线段相交交点)

    题目大意:有一个不反光并且不透光的管道,现在有一束光线从最左端进入,问能达到的最右端是多少,输出x坐标.   分析:刚开始做是直接枚举两个点然后和管道进行相交查询,不过这样做需要考虑的太多,细节不容易 ...

  7. POJ 1269 Intersecting Lines(线段相交,水题)

    id=1269" rel="nofollow">Intersecting Lines 大意:给你两条直线的坐标,推断两条直线是否共线.平行.相交.若相交.求出交点. ...

  8. POJ 1269 Intersecting Lines(直线相交判断,求交点)

    Intersecting Lines Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 8342   Accepted: 378 ...

  9. poj 1269 Intersecting Lines——叉积求直线交点坐标

    题目:http://poj.org/problem?id=1269 相关知识: 叉积求面积:https://www.cnblogs.com/xiexinxinlove/p/3708147.html什么 ...

随机推荐

  1. sublime text 3如何安装插件

    原博客地址:http://blog.csdn.net/weixin_40682842/article/details/78727266 我自己的部分操作如下: 学习Sublime Text扩展插件的安 ...

  2. java的finalize()方法与C++的析构函数

    ---<java编程思想> 读书笔记 --- 2017/3/15 读<java编程思想>读到初始化与清理一章,文中提及java的finalize()方法,联想到了C++的析构函 ...

  3. 布隆(Bloom)过滤器 JAVA实现

    前言 Bloom过滤器,通过将字符串映射为信息指纹从而节省了空间.Bloom过滤器的原理为,将一个字符串通过一定算法映射为八个Hash值,将八个Hash值对应位置的Bitset位进行填充.在进行校验的 ...

  4. sizeof与strlen的不同

    sizeof操作符的结果类型是size_t,它在头文件中typedef为unsigned int类型. 该类型保证能容纳实现所建立的最大对象的字节大小. sizeof是算符,strlen是函数. si ...

  5. console.log 用法

    转自http://www.cnblogs.com/ctriphire/p/4116207.html 大家都有用过各种类型的浏览器,每种浏览器都有自己的特色,本人拙见,在我用过的浏览器当中,我是最喜欢C ...

  6. 阿里巴巴Java开发手册评

    2016年底的时候阿里巴巴公开了其在内部使用的Java编程规范.随后进行了几次版本修订,目前的版本为v1.0.2版.下载地址可以在其官方社区-云栖社区https://yq.aliyun.com/art ...

  7. J.U.C JMM. pipeline.指令重排序,happen-before(续MESI协议)

    缓存(Cache)       CPU的读/写(以及取指令)单元正常情况下甚至都不能直接访问内存——这是物理结构决定的:CPU都没有管脚直接连到内存.相反,CPU和一级缓存(L1 Cache)通讯,而 ...

  8. maven 阿里云仓库配置

    <!-- 设定主仓库,按设定顺序进行查找. --> <repositories> <repository> <id>nexus-aliyun</i ...

  9. AI_深度学习概论

    什么是是神经网络? 假如有6间房屋的数据集,已知房子的面积,单位是平方米或平方英尺,已知房子的价格.如果通过这6间房子的价格和房子的面积,预测房子的价格,首先要建立起一个数据模型 ,x轴为价格,y轴为 ...

  10. Netty初探

    匠心零度 转载请注明原创出处,谢谢! 说在前面 为什么我们需要学习netty?谈谈自己的看法,由于本人水平有限,如果有那里不对,希望各位大佬积极指出,欢迎在留言区进行评论交流.探讨. 由于移动互联网的 ...