【POJ 1269】判断两直线相交】的更多相关文章

题目大意: t个测试用例 每次给出一对直线的两点 判断直线的相对关系 平行输出NODE 重合输出LINE 相交输出POINT和交点坐标 1.直线平行 两向量叉积为0 2.求两直线ab与cd交点 设直线ab上点为 a+(b-a)t,t为变量 交点需满足在直线cd上 则(d-c)*(a+t(b-a)-c)=0(外积) 分解为加减式 将t放在等号左边 其他放在右边 化简推导得t=(d-c)*(c-a)/(d-c)*(b-a) 则交点为a+(b-a)*((d-c)*(c-a)/(d-c)*(b-a))…
Intersecting Lines Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 8637   Accepted: 3915 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…
题 利用叉积解方程 #include <cstdio> #define MAX 1<<31 #define dd double int xmult(dd x1,dd y1,dd x2,dd y2,dd x,dd y){ return (x1-x)*(y2-y)-(x2-x)*(y1-y); } int main(){ int n; dd x1,y1,x2,y2,x3,y3,x4,y4; scanf("%d",&n); puts("INTERSE…
Pick-up sticks Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 11335   Accepted: 4250 Description Stan has n sticks of various length. He throws them one at a time on the floor in a random way. After finishing throwing, Stan tries to fin…
题目传送门:POJ 1269 Intersecting Lines 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…
http://poj.org/problem?id=1127   Description In the game of Jack Straws, a number of plastic or wooden "straws" are dumped on the table and players try to remove them one-by-one without disturbing the other straws. Here, we are only concerned wi…
Jack Straws Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 2911   Accepted: 1322 Description In the game of Jack Straws, a number of plastic or wooden "straws" are dumped on the table and players try to remove them one-by-one witho…
题目链接:http://poj.org/problem?id=1269 题面: 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) interse…
Intersecting Lines Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 13605   Accepted: 6049 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…
Pick-up sticks Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1872    Accepted Submission(s): 706 Problem Description Stan has n sticks of various length. He throws them one at a time on the fl…
题目链接 题意 判断两条直线的位置关系,重合/平行/相交(求交点). 直线以其上两点的形式给出(点坐标为整点). 思路 写出直线的一般式方程(用\(gcd\)化为最简), 计算\(\begin{vmatrix}a1&b1\\a2&b2\end{vmatrix}\), 若不为\(0\),则两直线有交点,\[x=\frac{\begin{vmatrix}-c1&b1\\-c2&b2\end{vmatrix}}{\begin{vmatrix}a1&b1\\a2&b…
Jack Straws In the game of Jack Straws, a number of plastic or wooden "straws" are dumped on the table and players try to remove them one-by-one without disturbing the other straws. Here, we are only concerned with if various pairs of straws are…
You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6837 Accepted Submission(s): 3303 Problem Description Many geometry(几何)problems were designed in the ACM/ICPC. A…
You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6425    Accepted Submission(s): 3099 Problem Description Many geometry(几何)problems were designed in the ACM/I…
最近需要用到矩形相交算法的简单应用,所以特地拿一个很简单的算法出来供新手参考,为什么说是给新手的参考呢因为这个算法效率并不是很高,但是这个算法只有简简单单的三行.程序使用了两种方法来判断是否重叠/相交,如果有兴趣可以看一下,如果觉得有bug可以留言.代码仅供参考. C#中矩形的方法为Rectangl(起始点坐标, 矩形的大小)或Rectangl(起始点x坐标, 起始点y坐标, 矩形宽, 矩形高),起始点为矩形区域的左上角. 方法一 姑且叫做“井字法”吧,延长其中一个矩形的四边使其形成一“井”字(…
题意:    判断直线间位置关系: 相交,平行,重合 include <iostream> #include <cstdio> using namespace std; struct Point { int x , y; Point(, ) :x(a), y(b) {} }; struct Line { Point s, e; int a, b, c;//a>=0 Line() {} Line(Point s1,Point e1) : s(s1), e(e1) {} void…
http://poj.org/problem?id=3304 Segments Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9449   Accepted: 2902 Description Given n segments in the two dimensional space, write a program, which determines if there exists a line such that a…
题目大意: 给定n条线段的端点 依次放上n条线段 判断最后在最上面(不被覆盖)的线段有哪些 到当前线段后 直接与之前保存的未被覆盖的线段判断是否相交就可以了 #include <cstdio> #include <algorithm> #include <string.h> #include <set> #include <cmath> #define INF 0x3f3f3f3f using namespace std; ; double ad…
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... 乱发文的同学真是该死...浪费我几个小时的生命... /********************* Template ************************/ #include <set> #include <map> #include <list> #include &l…
1840: Jack Straws  Time Limit(Common/Java):1000MS/10000MS     Memory Limit:65536KByteTotal Submit: 154            Accepted:119 Description In the game of Jack Straws, a number of plastic or wooden "straws" are dumped on the table and players try…
Pipe Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 8280   Accepted: 2483 Description The GX Light Pipeline Company started to prepare bent pipes for the new transgalactic light pipeline. During the design phase of the new pipe shape th…
链接: http://poj.org/problem?id=1039 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22013#problem/B Pipe Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 8350   Accepted: 2501 Description The GX Light Pipeline Company started to prep…
题目描述 Two out-of-control cars crashed within about a half-hour Wednesday afternoon on Deer Park Avenue. This accident alarmed the district government.It jumpstarted a vibrant new technology to predict potential car accidents.Engineers depicted a movin…
一 题面 POJ1127 二 分析 在平面几何中,判断两线段相交的方法一般是使用跨立实验.但是这题考虑了非严格相交,即如何两个线段刚好端点相交则也是相交的,所以还需要使用快速排斥实验. 这里参考并引用了TangMoon 博客. 1.快速排斥实验 由于两个点作为矩形的两个斜对角线端点可以确定一个矩形,则根据两个点确定一个向量,两个向量显然可以确定两个矩形. 对于快速排斥实验,也可尝试逆向思维,如果判断让两个向量确定的两个矩形是否相交部分,相当于判断两个矩形没有相交部分的反. 具体条件就是a.其中一…
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, p2, p3, p4,直线L1,L2分别穿过前两个和后两个点.来判断直线L1和L2的关系 这三种关系一个一个来看: 1. 共线. 如果两条直线共线的话,那么另外一条直线上的点一定在这一条直线上.所以p3在p1p2上,所以用get_direction(p1, p2, p3)来判断p3相对于p1p2的关…
Intersecting Lines Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 8342   Accepted: 3789 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…
题目链接:POJ 1269 Problem 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 becau…
// 判断线段和直线相交 POJ 3304 // 思路: // 如果存在一条直线和所有线段相交,那么平移该直线一定可以经过线段上任意两个点,并且和所有线段相交. #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <map> #include <set> #include <queue> #includ…
// 直线相交 POJ 1269 // #include <bits/stdc++.h> #include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <math.h> using namespace std; #define LL long long typedef pair<int,int> pii; con…
原题 在金字塔内有一个宝藏p(x,y),现在要取出这个宝藏. 在金字塔内有许多墙,为了进入宝藏所在的房间必须把墙炸开,但是炸墙只能炸每个房间墙的中点. 求将宝藏运出城堡所需要的最小炸墙数. 判断点和直线相交. 枚举每道墙的两个端点和p的连线这条线段和墙的交点的次数最小值即为需要炸墙的最小次数. [注意当墙数为零时输出1:] #include<cstdio> #include<algorithm> #define N 33 using namespace std; int ans=0…