题目传送门

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

  1. 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 ...

  2. 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, ...

  3. 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 ...

  4. 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 ...

  5. 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 ...

  6. 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) ...

  7. 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 ...

  8. 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 ...

  9. 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

  1. 你真的会用AsyncTask吗?(一)

    一个典型AsyncTask的. view source print? 01 public class DialogTestActivity extends Activity { 02     priv ...

  2. MFC 最大化 的时候控件 按比例变大

    在dlg类头文件中声明CPoint Old; 在BEGIN_MESSAGE_MAP()和END_MESSAGE_MAP()声明一个映射:ON_WM_SIZE() 这样以后就可以在M_SIZE事件的时候 ...

  3. java对象与json对象间的相互转换

    工程中所需的jar包,因为在网上不太好找,所以我将它放到我的网盘里了,如有需要随便下载. 点击下载 1.简单的解析json字符串 首先将json字符串转换为json对象,然后再解析json对象,过程如 ...

  4. [另开新坑] 算导v3 #26 最大流 翻译

    26 最大流 就像我们可以对一个路网构建一个有向图求最短路一样,我们也可以将一个有向图看成是一个"流量网络(flow network)",用它来回答关于流的问题. Just as ...

  5. Python字符串与数字互转,数字格式化

    # -*- coding: gbk -*- import re #将数字格式化为带三位数逗号的字符串 def formatNumber(number): numStr='%d'%number form ...

  6. 【Docker】来自官方映像的 6 个 Dockerfile 技巧

    本文将根据我从官方镜像学到的经验,讲解编写Dockerfile的技巧.   1. 选择Debian  官方镜像的大多数Dockerfile,不管是直接还是通过其他镜像,都是基于Debian的.Dock ...

  7. Repeated DNA Sequences

    All DNA is composed of a series of nucleotides abbreviated as A, C, G, and T, for example: "ACG ...

  8. 对于sharepoint 的解决方案的实际说明

    对于sharepoint 的解决方案  实际上就是cab的包 你把***.wsp改为***.cab我们就可以查看这个包中的所有内容了

  9. Java for LeetCode 165 Compare Version Numbers

    Compare two version numbers version1 and version2.If version1 > version2 return 1, if version1 &l ...

  10. Java for LeetCode 047 Permutations II

    Given a collection of numbers that might contain duplicates, return all possible unique permutations ...