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

我会说这种水题我手推公式+码代码用了1.5h?

还好新的一年里1A了~~~~

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mkpii make_pair<int, int>
#define pdi pair<double, int>
#define mkpdi make_pair<double, int>
#define pli pair<ll, int>
#define mkpli make_pair<ll, int>
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << (#x) << " = " << (x) << endl
#define error(x) (!(x)?puts("error"):0)
#define printarr2(a, b, c) for1(_, 1, b) { for1(__, 1, c) cout << a[_][__]; cout << endl; }
#define printarr1(a, b) for1(_, 1, b) cout << a[_] << '\t'; cout << endl
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; } const double eps=1e-6;
struct Pt { double x, y; Pt(double _x=0, double _y=0) : x(_x), y(_y) {} };
int dcmp(double a) { if(abs(a)<eps) return 0; return a<0?-1:1; }
typedef Pt Vt;
Vt operator+ (const Pt &a, const Pt &b) { return Vt(a.x+b.x, a.y+b.y); }
Vt operator- (const Pt &a, const Pt &b) { return Vt(a.x-b.x, a.y-b.y); }
Vt operator* (const Pt &a, const double &b) { return Vt(a.x*b, a.y*b); }
bool operator== (const Pt &a, const Pt &b) { return dcmp(a.x-b.x)==0 && dcmp(a.y-b.y)==0; }
double Cross(Vt a, Vt b) { return a.x*b.y-b.x*a.y; } struct Line {
Pt p; Vt v;
Line() {}
Line(Pt &a, Pt &b) { p=a; v=b-a; }
}; Pt getLLP(Line &a, Line &b) {
static Pt p, q;
static Vt u, w, v;
p=a.p; q=b.p;
v=a.v; w=b.v;
u=p-q;
double t1=Cross(w, u)/Cross(v, w);
return p+v*t1;
}
// -1:xiangjiao 0:chonghe 1:pingxing
int LineAndLine(Line &p, Line &q) {
if(dcmp(Cross(p.v, q.v))!=0) return -1;
return dcmp(Cross(q.p-p.p, q.v))==0 && dcmp(Cross(q.p-p.p, p.v))==0;
}
int main() {
int n;
while(~scanf("%d", &n)) {
puts("INTERSECTING LINES OUTPUT");
Line l[2]; Pt p[4];
while(n--) {
rep(k, 4) scanf("%lf%lf", &p[k].x, &p[k].y);
l[0]=Line(p[0], p[1]);
l[1]=Line(p[2], p[3]);
int c=LineAndLine(l[0], l[1]);
if(c==-1) { Pt pt=getLLP(l[0], l[1]); printf("POINT %.2f %.2f\n", pt.x, pt.y); }
else if(c==0) puts("NONE");
else puts("LINE");
}
puts("END OF OUTPUT");
}
return 0;
}

  


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. 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. 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 ...

  3. poj 1269 Intersecting Lines

    题目链接:http://poj.org/problem?id=1269 题目大意:给出四个点的坐标x1,y1,x2,y2,x3,y3,x4,y4,前两个形成一条直线,后两个坐标形成一条直线.然后问你是 ...

  4. 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, ...

  5. ●POJ 1269 Intersecting Lines

    题链: http://poj.org/problem?id=1269 题解: 计算几何,直线交点 模板题,试了一下直线的向量参数方程求交点的方法. (方法详见<算法竞赛入门经典——训练指南> ...

  6. POJ 1269 - Intersecting Lines - [平面几何模板题]

    题目链接:http://poj.org/problem?id=1269 Time Limit: 1000MS Memory Limit: 10000K Description We all know ...

  7. 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 ...

  8. POJ 1269 Intersecting Lines(几何)

    题目链接 题意 : 给你两条线段的起点和终点,一共四个点,让你求交点坐标,如果这四个点是共线的,输出“LINE”,如果是平行的就输出“NONE”. 思路 : 照着ZN留下的模板果然好用,直接套上模板了 ...

  9. 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. ...

  10. POJ 1269 Intersecting Lines --计算几何

    题意: 二维平面,给两条线段,判断形成的直线是否重合,或是相交于一点,或是不相交. 解法: 简单几何. 重合: 叉积为0,且一条线段的一个端点到另一条直线的距离为0 不相交: 不满足重合的情况下叉积为 ...

随机推荐

  1. poj2253 最短路 floyd Frogger

    Frogger Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 28825   Accepted: 9359 Descript ...

  2. Nginx图片剪裁模块探究 http_image_filter_module

    官方地址:http://nginx.org/en/docs/http/ngx_http_image_filter_module.html 煮酒品茶:前半部安装和官方说明,后半部分实践 #yum ins ...

  3. object-c面向对象1

    ---恢复内容开始--- 类,对象,方法,属性. 类是object-c一种重要的数据类型,是组成object-c程序的基本要素.object-c的类声明和实现包括两个部分:接口和实现部分. @inte ...

  4. Apache配置文件中的deny和allow的使用

    Apache配置文件中的deny和allow的使用 由于产品的需要,最近在配置apache的负载均衡功能,但是在配置虚拟主机的访问权限的时候我们遇到了一些问题.主要问题是deny和allow的执行顺序 ...

  5. java常用算法

    冒泡排序: //降序 public static int[] bubbleSort(int[] array){ for(int i = 0; i < array.length; i++){ in ...

  6. 转mysql存储引擎memory,ndb,innodb之选择

    1 mysql的innodb和cluster的NDB引擎都支持事务,在有共同的特性外,也有不同之处:以mysql cluster NDB 7.3和MySQL 5.6之InnoDB为例:ndb7.3基于 ...

  7. 解决虚拟机 正在决定eht0 的ip信息失败 无链接-- 添加虚拟网卡

    添加步骤:1.进入设备管理器 2.点下一步3.继续下一步 4.继续往下走

  8. C++动态内存管理之shared_ptr、unique_ptr

    C++中的动态内存管理是通过new和delete两个操作符来完成的.new操作符,为对象分配内存并调用对象所属类的构造函数,返回一个指向该对象的指针.delete调用时,销毁对象,并释放对象所在的内存 ...

  9. Android之Activity框架

    在安卓应用中,经常需要Activity中经常需要有大量相似的Activity类,这些类往往有相似的结构与功能,因此产生了大量重复代码,为此,以下提供一种方法有效的降低了代码冗余. 定义Activity ...

  10. Hadoop配置文件

    部分内容参考:http://www.linuxqq.net/archives/964.html  http://slaytanic.blog.51cto.com/2057708/1100974/ ht ...