/************************************************************************/ /*线段与WGS84椭球求交 x^2/a^2+y^2/a^2+z^2/b^2=1 (x-x0)/m=(y-y0)/n=(z-z0)/p=t m=x1-x0 n=y1-y0 p=z1-z0 p0线段起始点 p1线段终点 center椭球球心 a = osg::WGS_84_RADIUS_EQUATOR;//长轴 b = osg::WGS_84_RAD

链接 几何细节题. 对于每一个障碍物可以求出它在地产线上的覆盖区间,如下图. 紫色部分即为每个障碍物所覆盖掉的区间,求出所有的,扫描一遍即可. 几个需要注意的地方:直线可能与地产线没有交点,可视区间可能包含地产线的端点,扫描的时候保留当前扫到的最大值. 代码中的数据很经典,供参考. #include <iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<st

// 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

// 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

http://poj.org/problem?id=1556 The Doors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 6120   Accepted: 2455 Description You are to find the length of the shortest path through a chamber containing obstructing walls. The chamber will a

题目: 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

题意: 给出n条线段两个端点的坐标,问所有线段投影到一条直线上,如果这些所有投影至少相交于一点就输出Yes!,否则输出No!. 思路: 计算几何.这道题要思考到两点: 1:把问题转化为是否存在一条直线与每条线段都有交点.证明:若存在一条直线l和所有线段相交,作一条直线m和l垂直,则m就是题中要求的直线,所有线段投影的一个公共点即为垂足. 2:枚举两两线段的各一个端点,连一条直线,再判断剩下的线段是否都和这条直线有交点.证明:若有l和所有线段相交,则可保持l和所有线段相交,左右平移l到和某一线段交

Jack Straws 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 with if various pairs o

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

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

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

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)=0 (p3-p0)X(p2-p0)=0 展开后即是 (y1-y2)x0+(x2-x1)y0+x1y2-x2y1=0 (y3-y4)x0+(x4-x3)y0+x3y4-x4y3=0 将x0,y0作为变量求解二元一次方程组. 假设有二元一次方程组 a1x+b1y+c1=0; a2x+b2y+c2=0

7.3 Given two lines on a Cartesian plane, determine whether the two lines would intersect. 这道题说是在笛卡尔坐标系中,让我们确定两条直线是否相交. 那么我们首先要写个直线的类来表示直线,最常见的表示方法为y=kx+b,k为斜率,b为与y轴的交点,然后我们来考虑什么情况下两条直线会相交,首先,如果两条直线完全重合的话,应该也算是相交的,其次如果两条直线不平行的话,那么也是相交的,那么我们判断相交条件主要基于

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

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

求相交点 /* 线段相交模板:判相交.求交点 */ #include<stdio.h> #include<string.h> #include<stdlib.h> #include<math.h> ; struct Point{ double x,y; }; Point P_ans; double cross( Point a,Point b,Point c ){ return ( b.x-a.x )*( c.y-a.y )-( b.y-a.y )*( c.

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

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

Segments Time Limit: 1000MS   Memory Limit: 65536K       Description Given n segments in the two dimensional space, write a program, which determines if there exists a line such that after projecting these segments on it, all projected segments have