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. MYSQL数据库增量备份

    MySQL数据库增量备份,在这之前修改我们的数据库配置文件/etc/my.cnf开启bin-log日志功能即可.接下来是我参考了下网上的一些方法,自己写的,主要还是要能学到他的一些思路和方法. #fu ...

  2. centos7+cdh5.10.0搭建

    一.选择环境: 1.说明 本次部署使用台机器,3台用于搭建CDH集群,1台为内部源.内部源机器是可以连接公网的,可以提前部署好内部源,本次部署涉及到的服务器的hosts配置如下: 192.168.10 ...

  3. redis3.0 集群在windows上的配置(转)

    1. 安装Redis版本:win-3.0.501https://github.com/MSOpenTech/redis/releases页面有,我下载的是zip版本的:Redis-x64-3.0.50 ...

  4. JAVA开发中遇到的异常总结

    最常见的五种异常:必会,面试题: 算术异常类:ArithmeticExecption 空指针异常类:NullPointerException 类型强制转换异常:ClassCastException 数 ...

  5. Django App(二) Connect Mysql & defualt App admin

    这一篇接着上一篇polls App自动创建admin app.     1.安装数据库 这里的内容从官网看越看越像 EntityFramework的内容.Python支持SQLite,MySql,Or ...

  6. 安装Ruby、Sass与Compass

    sass基于Ruby语言开发而成,因此安装sass前需要安装Ruby.(注:mac下自带Ruby无需在安装Ruby!) window下安装SASS首先需要安装Ruby,先从官网下载Ruby并安装.安装 ...

  7. Nginx的启动(start),停止(stop)命令

    http://blog.csdn.net/u010739551/article/details/51654859 查看Nginx的版本号:nginx -V 启动Nginx:start nginx 快速 ...

  8. 邓_PPT

    如何拯救一份丑到爆的PPT? "小邓,这是我做的PPT,你优化优化,明天早上给我,上午客户来要用." 领导,你这是PPT嘛,明明就是word嘛. "小张啊,你看看我这个P ...

  9. ap web

    apapplication端吧    应用程序端   也C-S架构Cweb网页端   般封装httpservletrequest和httpservletresponse对象处理些操作  b-s架构

  10. 用Express、MySQL搭建项目(接口以及静态文件获取、文件上传等)

    一.简介 本文将主要基于node.js使用express框架搭建一个后台环境,包括如何自定义项目目录.所用依赖以及中间件.路由以及模板引擎.接口数据获取以及文件上传等内容. 二.后台环境搭建 1.新建 ...