描述


http://poj.org/problem?id=1269

给出两条直线,判断它们是平行,重合,还是相交,如果相交,求出交点.

分析


比较裸的一道题.学习了直线的写法(参数方程)

 #include <cstdio>
#include <cmath>
using namespace std; const double eps=1e-; struct pt{ double x,y; pt(double x=,double y=):x(x),y(y){} };
typedef pt vt;
int dcmp(double x){ if(fabs(x)<eps) return ; return x>?:-; }
vt operator + (vt a,vt b){ return vt(a.x+b.x,a.y+b.y); }
vt operator - (vt a,vt b){ return vt(a.x-b.x,a.y-b.y); }
vt operator * (vt a,double p){ return vt(a.x*p,a.y*p); }
double cross(vt a,vt b){ return a.x*b.y-a.y*b.x; }
struct line{
pt p; vt v;
line(){}
line(pt a,pt b){ p=a; v=b-a; }
};
int line_intersection(line A,line B){
if(dcmp(cross(A.v,B.v)!=)) return -;
return dcmp(cross(A.v,A.p-B.p))==;
}
pt get_line_intersection(line A,line B){
vt v=A.v,w=B.v,u=A.p-B.p;
double t=cross(w,u)/cross(v,w);
return A.p+v*t;
}
int main(){
int n;
scanf("%d",&n);
puts("INTERSECTING LINES OUTPUT");
while(n--){
pt p[]; line l[];
for(int i=;i<;i++) scanf("%lf%lf",&p[i].x,&p[i].y);
l[]=line(p[],p[]);
l[]=line(p[],p[]);
int t=line_intersection(l[],l[]);
if(t==-){
pt x=get_line_intersection(l[],l[]);
printf("POINT %.2lf %.2lf\n",x.x,x.y);
}
else if(t==) puts("LINE");
else puts("NONE");
}
puts("END OF OUTPUT");
return ;
}
Intersecting Lines
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 13622   Accepted: 6060

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

POJ_1269_Intersecting_Lines_(计算几何基础)的更多相关文章

  1. nyis oj 68 三点顺序 (计算几何基础)

    三点顺序 时间限制:1000 ms  |  内存限制:65535 KB 难度:3 描写叙述 如今给你不共线的三个点A,B,C的坐标,它们一定能组成一个三角形,如今让你推断A,B,C是顺时针给出的还是逆 ...

  2. 【BZOJ】1043: [HAOI2008]下落的圆盘(计算几何基础+贪心)

    http://www.lydsy.com/JudgeOnline/problem.php?id=1043 唯一让我不会的就是怎么求圆的周长并QAAQ... 然后发现好神!我们可以将圆弧变成$[0, 2 ...

  3. 计算几何基础——矢量和叉积 && 叉积、线段相交判断、凸包(转载)

    转载自 http://blog.csdn.net/william001zs/article/details/6213485 矢量 如果一条线段的端点是有次序之分的话,那么这种线段就称为 有向线段,如果 ...

  4. BZOJ_1610_[Usaco2008_Feb]_Line连线游戏_(计算几何基础+暴力)

    描述 http://www.lydsy.com/JudgeOnline/problem.php?id=1610 给出n个点,问两两确定的直线中,斜率不同的共有多少条. 分析 暴力枚举直线,算出来斜率放 ...

  5. 二维计算几何基础题目泛做(SYX第一轮)

    题目1: POJ 2318 TOYS 题目大意: 给一个有n个挡板的盒子,从左到右空格编号为0...n.有好多玩具,问每个玩具在哪个空格里面. 算法讨论: 直接叉积判断就可以.注意在盒子的边界上面也算 ...

  6. 计算几何基础算法几何C++实现

    This file is implementation of Common Common Computational Geometry Algorithms.Please please pay att ...

  7. 【POJ】1556 The Doors(计算几何基础+spfa)

    http://poj.org/problem?id=1556 首先路径的每条线段一定是端点之间的连线.证明?这是个坑...反正我是随便画了一下图然后就写了.. 然后re是什么节奏?我记得我开够了啊.. ...

  8. 【POJ】2318 TOYS(计算几何基础+暴力)

    http://poj.org/problem?id=2318 第一次完全是$O(n^2)$的暴力为什么被卡了-QAQ(一定是常数太大了...) 后来排序了下点然后单调搞了搞..(然而还是可以随便造出让 ...

  9. 【POJ】2653 Pick-up sticks(计算几何基础+暴力)

    http://poj.org/problem?id=2653 我很好奇为什么这样$O(n^2)$的暴力能过.... 虽然说这是加了链表优化的,但是最坏不也是$O(n^2)$吗...(只能说数据太弱.. ...

随机推荐

  1. C++ txt文档读取

    void readfile(string filepath){ ifstream myfile; if (!myfile) { cout << "打开文件出错!"; e ...

  2. 初学C++,开博第一篇

    几个维度相同的数组转置算法,这两种完全相同://注意:如果维度不同,转置会出错误,因为下标会越界...解决办法是把数组的维度调齐,或者是写到另一新数组中. for(int i=0;i<row;i ...

  3. 393. UTF-8 Validation

    393. UTF-8 Validation 这个题很明确,刚开始我以为只能是一个utf,长度大于5的都判断为false,后来才明白题意. 有个小trick,就是长度大于1的时候,判断第一个数字开始1的 ...

  4. Build Error 6041: Internal build error

    Note: Following content is reprinted from the Original article Flexera : Build Error 6041. Only for ...

  5. 【转】提高PHP性能的53个技巧

    PHP技巧汇总:提高PHP性能的53个技巧用单引号代替双引号来包含字符串,这样做会更快一些.因为PHP会在双引号包围的字符串中搜寻变量,单引号则不会,注意:只有echo能这么做,它是一种可以把多个字符 ...

  6. ssh连接失败解决方法

    执行如下命令: ssh-keygen -t dsa -P '' -f /etc/ssh/ssh_host_dsa_key ssh-keygen -t rsa -P '' -f /etc/ssh/ssh ...

  7. Cloudcraft: 云架构图形可视化(智能AWS图表)

    Cloudcraft: 云架构图形可视化(智能AWS图表) 2016.09.11 官方网站: https://cloudcraft.co/ Cloudcraft是一个Web应用,用图形表示各种AWS服 ...

  8. Swift 学习笔记1

    最近在看Swift,努力在看相关的文档以及书籍,因为Swift3.0的更新,以及它开源了,所以打算写一些关于Swift的相关文章.让Swift能够更好的被我理解

  9. 使用phpize安装php模块

    1.下载包 2./usr/local/php/bin/phpize 3../configure --enable-soap  --with-php-config=/usr/local/php/bin/ ...

  10. 在SQL中导入Excel数据时强制以文本类型导入

    Excel不是关系型数据库,在导入到sql中时对于数值型,sql有时int型会处理成float,有时数字文本混排的列,sql会认为是数值型,导入的结果有的数据变成了null,但是用sql导出excel ...