简单几何(直线位置) POJ 1269 Intersecting Lines
题意:判断两条直线的位置关系,共线或平行或相交
分析:先判断平行还是共线,最后就是相交。平行用叉积判断向量,共线的话也用叉积判断点,相交求交点
/************************************************
* Author :Running_Time
* Created Time :2015/10/24 星期六 09:08:55
* File Name :POJ_1269.cpp
************************************************/ #include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std; #define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
struct Point { //点的定义
double x, y;
Point (double x=0, double y=0) : x (x), y (y) {}
};
typedef Point Vector; //向量的定义
Point read_point(void) { //点的读入
double x, y;
scanf ("%lf%lf", &x, &y);
return Point (x, y);
}
double polar_angle(Vector A) { //向量极角
return atan2 (A.y, A.x);
}
double dot(Vector A, Vector B) { //向量点积
return A.x * B.x + A.y * B.y;
}
double cross(Vector A, Vector B) { //向量叉积
return A.x * B.y - A.y * B.x;
}
int dcmp(double x) { //三态函数,减少精度问题
if (fabs (x) < EPS) return 0;
else return x < 0 ? -1 : 1;
}
Vector operator + (Vector A, Vector B) { //向量加法
return Vector (A.x + B.x, A.y + B.y);
}
Vector operator - (Vector A, Vector B) { //向量减法
return Vector (A.x - B.x, A.y - B.y);
}
Vector operator * (Vector A, double p) { //向量乘以标量
return Vector (A.x * p, A.y * p);
}
Vector operator / (Vector A, double p) { //向量除以标量
return Vector (A.x / p, A.y / p);
}
bool operator < (const Point &a, const Point &b) { //点的坐标排序
return a.x < b.x || (a.x == b.x && a.y < b.y);
}
bool operator == (const Point &a, const Point &b) { //判断同一个点
return dcmp (a.x - b.x) == 0 && dcmp (a.y - b.y) == 0;
}
double length(Vector A) { //向量长度,点积
return sqrt (dot (A, A));
}
double angle(Vector A, Vector B) { //向量转角,逆时针,点积
return acos (dot (A, B) / length (A) / length (B));
}
double area_triangle(Point a, Point b, Point c) { //三角形面积,叉积
return fabs (cross (b - a, c - a)) / 2.0;
}
Vector rotate(Vector A, double rad) { //向量旋转,逆时针
return Vector (A.x * cos (rad) - A.y * sin (rad), A.x * sin (rad) + A.y * cos (rad));
}
Vector nomal(Vector A) { //向量的单位法向量
double len = length (A);
return Vector (-A.y / len, A.x / len);
}
Point point_inter(Point p, Vector V, Point q, Vector W) { //两直线交点,参数方程
Vector U = p - q;
double t = cross (W, U) / cross (V, W);
return p + V * t;
}
double dis_to_line(Point p, Point a, Point b) { //点到直线的距离,两点式
Vector V1 = b - a, V2 = p - a;
return fabs (cross (V1, V2)) / length (V1);
}
double dis_to_seg(Point p, Point a, Point b) { //点到线段的距离,两点式 if (a == b) return length (p - a);
Vector V1 = b - a, V2 = p - a, V3 = p - b;
if (dcmp (dot (V1, V2)) < 0) return length (V2);
else if (dcmp (dot (V1, V3)) > 0) return length (V3);
else return fabs (cross (V1, V2)) / length (V1);
}
Point point_proj(Point p, Point a, Point b) { //点在直线上的投影,两点式
Vector V = b - a;
return a + V * (dot (V, p - a) / dot (V, V));
}
bool inter(Point a1, Point a2, Point b1, Point b2) { //判断线段相交,两点式
double c1 = cross (a2 - a1, b1 - a1), c2 = cross (a2 - a1, b2 - a1),
c3 = cross (b2 - b1, a1 - b1), c4 = cross (b2 - b1, a2 - b1);
return dcmp (c1) * dcmp (c2) < 0 && dcmp (c3) * dcmp (c4) < 0;
}
bool on_seg(Point p, Point a1, Point a2) { //判断点在线段上,两点式
return dcmp (cross (a1 - p, a2 - p)) == 0 && dcmp (dot (a1 - p, a2 - p)) < 0;
}
double area_poly(Point *p, int n) { //多边形面积
double ret = 0;
for (int i=1; i<n-1; ++i) {
ret += fabs (cross (p[i] - p[0], p[i+1] - p[0]));
}
return ret / 2;
} int main(void) {
int T; scanf ("%d", &T);
Point a1, a2, b1, b2;
puts ("INTERSECTING LINES OUTPUT");
while (T--) {
a1 = read_point ();
a2 = read_point ();
b1 = read_point ();
b2 = read_point ();
Vector A = a2 - a1, B = b2 - b1;
if (cross (A, B) == 0) {
if (cross (b1 - a1, b2 - a1) == 0 && cross (b1 - a2, b2 - a2) == 0) puts ("LINE");
else puts ("NONE");
}
else {
Point ans = point_inter (a1, a2 - a1, b1, b2 - b1);
printf ("POINT %.2f %.2f\n", ans.x, ans.y);
}
}
puts ("END OF OUTPUT"); return 0;
}
简单几何(直线位置) POJ 1269 Intersecting Lines的更多相关文章
- POJ 1269 Intersecting Lines(判断两直线位置关系)
题目传送门:POJ 1269 Intersecting Lines Description We all know that a pair of distinct points on a plane ...
- 判断两条直线的位置关系 POJ 1269 Intersecting Lines
两条直线可能有三种关系:1.共线 2.平行(不包括共线) 3.相交. 那给定两条直线怎么判断他们的位置关系呢.还是用到向量的叉积 例题:POJ 1269 题意:这道题是给定四个点p1, ...
- POJ 1269 Intersecting Lines (判断直线位置关系)
题目链接:POJ 1269 Problem Description We all know that a pair of distinct points on a plane defines a li ...
- 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 ...
- POJ 1269 - Intersecting Lines 直线与直线相交
题意: 判断直线间位置关系: 相交,平行,重合 include <iostream> #include <cstdio> using namespace std; str ...
- POJ 1269 Intersecting Lines【判断直线相交】
题意:给两条直线,判断相交,重合或者平行 思路:判断重合可以用叉积,平行用斜率,其他情况即为相交. 求交点: 这里也用到叉积的原理.假设交点为p0(x0,y0).则有: (p1-p0)X(p2-p0) ...
- 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 ...
- POJ 1269 Intersecting Lines(直线求交点)
Description We all know that a pair of distinct points on a plane defines a line and that a pair of ...
- POJ 1269 Intersecting Lines 直线交
不知道谁转的计算几何题集里面有这个题...标题还写的是基本线段求交... 结果题都没看就直接敲了个线段交...各种姿势WA一遍以后发现题意根本不是线段交而是直线交...白改了那个模板... 乱发文的同 ...
随机推荐
- 调用{dede:likewords}为dedecms添加相关搜索词
经常看到一些大型的网站会设置相关搜索,即使访客搜索的内容在本站暂时没有,它们也会展示一些其他搜索关键词,引导用户去点击查看,增加pv,提高用户体验:如果没有这些相关搜索,游客没有找到自己想要的内容就直 ...
- linux expect 简单讲解
来自http://blog.csdn.net/winstary/archive/2009/08/08/4422156.aspx使用expect实现自动登录的脚本,网上有很多,可是都没有一个明白的说明, ...
- HDOJ 1869
#include<stdio.h> ][]; #define inf 0xffffff; void floyed(int n) { int i,j,k; ;k<n;k++) { ;i ...
- microsoft office安装选择
office分为零售版和批量授权版 零售版(文件名以cn开头)需要提供序列号才可以安装,而批量授权版(文件名以SW开头)可以先安装试用一段时间.
- catalog、scheme区别
按照SQL标准的解释,在SQL环境下Catalog和Schema都属于抽象概念,可以把它们理解为一个容器或者数据库对象命名空间中的一个层次,主要用来解决命名冲突问题.从概念上说,一个数据库系统包含多个 ...
- Top K Frequent Elements
Given a non-empty array of integers, return the k most frequent elements. For example,Given [1,1,1,2 ...
- cocos2dx阴影层的实现
效果图 //ShadowLayer.h class ShadowLayer : public CCLayer { protected: ShadowLayer() :m_pRender(NULL) , ...
- mysql 我的学习
安装要求 安装环境:CentOS-6.3安装方式:源码编译安装 软件名称:mysql-cluster-gpl-7.2.6-linux2.6-x86_64.tar.gz下载地址:http://mysql ...
- cmd的rd命令简单解析
我们知道在Windows下cmd命令行中"rd 文件夹名称"可以删除空目录,"del 文件名"可以删除文件,那么怎么删除一个非空文件夹呢,命令如下: 比如删除文 ...
- mac OS X 安装svn
因为从10.5版本开始适用Mac OS,SVN一直都是默认安装的软件. 后来发现一个简单的办法. 如果你有安装XCode,只需要在code > Preferences > download ...