poj 1269 线段相交/平行

模板题注意原题中说的线段其实要当成没有端点的直线.被坑了= = #include <cmath> #include <cstdio> #include <iostream> #include <cstring> using namespace std; #define eps 1e-8 #define PI acos(-1.0)//3.14159265358979323846 //判断一个数是否为0,是则返回true,否则返回false #define z…

poj 1269 线段与线段相交

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…

poj 1066 线段相交

链接:http://poj.org/problem?id=1066 Treasure Hunt Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 5431 Accepted: 2246 Description Archeologists from the Antiquities and Curios Museum (ACM) have flown to Egypt to examine the great pyramid…

poj 2653 线段相交

题意:一堆线段依次放在桌子上,上面的线段会压住下面的线段,求找出没被压住的线段. sol:从下向上找,如果发现上面的线段与下面的相交,说明被压住了.break掉其实这是个n^2的算法,但是题目已经说了没被压住的线段不超过1000个,所以不会爆 #include<math.h> #include <stdio.h> #include <string.h> ]; int n; double X1,X2,Y1,Y2; #define eps 1e-8 #define PI…

poj 2653 线段相交裸题（解题报告）

#include<stdio.h> #include<math.h> const double eps=1e-8; int n; int cmp(double x) { if(fabs(x)<=eps)return 0; if(x<0)return -1; return 1; } struct Point { double x,y; Point (){} Point (double _x,double _y) { x=_x; y=_y; } Point operator…

Pipe - POJ 1039(线段相交交点)

题目大意:有一个不反光并且不透光的管道,现在有一束光线从最左端进入,问能达到的最右端是多少,输出x坐标. 分析:刚开始做是直接枚举两个点然后和管道进行相交查询,不过这样做需要考虑的太多,细节不容易掌控.后来发现其实只需要对接口进行一下相交查询就简单多了,因为只需要考虑能不能通过每个截口就可以了,而且这样做的好处还有没有平行线和重叠线的情况,因为所有的截口都是垂直于x轴的,换一种想法海阔太空啊. 代码如下: =========================================…

poj 1410 线段相交判断

http://poj.org/problem?id=1410 Intersection Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 11329 Accepted: 2978 Description You are to write a program that has to decide whether a given line segment intersects a given rectangle. An ex…

POJ 1269 (直线相交) Intersecting Lines

水题,以前总结的模板还是很好用的. #include <cstdio> #include <cmath> using namespace std; ; int dcmp(double x) { ; ? - : ; } struct Point { double x, y; Point(, ):x(x), y(y) {} }; typedef Point Vector; Point read_point() { double x, y; scanf("%lf%lf"…

Pick-up sticks - POJ 2653 (线段相交)

题目大意:有一个木棒,按照顺序摆放,求出去上面没有被别的木棍压着的木棍..... 分析:可以维护一个队列,如果木棍没有被压着就入队列,如果判断被压着,就让那个压着的出队列,最后把这个木棍放进队列,不过速度并不快,枚举才是最快的......据说是任意时刻没有超过1000个top sticks.....很难注意到. 代码如下: ===================================================================================…

The Doors - POJ 1556 (线段相交)

题目大意:有一个房间(左上角(0,10),右下角(10,0)),然后房间里有N面墙,每面墙上都有两个门,求出来从初始点(0,5),到达终点(10,5)的最短距离. 分析:很明显根据两点之间直线最短,所以所走的路线一定是点之间的连线,只需要判断一下这两点间知否有墙即可. 代码如下: =========================================================================================================…

POJ 2074 | 线段相交

#include<cstdio> #include<algorithm> #include<cstring> #include<cmath> #define eps 1e-8 using namespace std; bool dcmp(double x,double y) { if (fabs(x-y)>eps) return 1; return 0; } struct point { double x,y; point () {}; point (…

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. 思路:线段相交推断.求交点的水题.没什么好说的. struct Point{ double x, y; } ; struct Line{ Point a, b; } A, B; double xmult(Point p1, Point p2, Point p) { return (p1.x-p.x)*(p2…

POJ 2826 An Easy Problem? 判断线段相交

POJ 2826 An Easy Problem?! -- 思路来自kuangbin博客下面三种情况比较特殊,特别是第三种 G++怎么交都是WA,同样的代码C++A了 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; const double eps = 1e-8;…

poj 3304线段与直线相交

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…

POJ 1039 Pipe（直线和线段相交判断，求交点）

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…

POJ 1066 Treasure Hunt(线段相交判断)

Treasure Hunt Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 4797 Accepted: 1998 Description Archeologists from the Antiquities and Curios Museum (ACM) have flown to Egypt to examine the great pyramid of Key-Ops. Using state-of-the-ar…

线段相交 poj 1066

// 线段相交 poj 1066 // 思路:直接枚举每个端点和终点连成线段,判断和剩下的线段相交个数 // #include <bits/stdc++.h> #include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <vector> #include <math.h> using namespace std…

线段相交 POJ 2653

// 线段相交 POJ 2653 // 思路:数据比较水,据说n^2也可以过 // 我是每次枚举线段,和最上面的线段比较 // O(n*m) // #include <bits/stdc++.h> #include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <vector> #include <math.h>…

直线相交 POJ 1269

// 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…

poj 3304(直线与线段相交)

传送门:Segments 题意:线段在一个直线上的摄影相交求求是否存在一条直线,使所有线段到这条直线的投影至少有一个交点分析:可以在共同投影处作原直线的垂线,则该垂线与所有线段都相交<==> 是否存在一条直线与所有线段都相交. 去盗了一份bin神的模板,用起来太方便了... #include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #incl…

[poj 1127]Jack Straws[线段相交][并查集]

题意: 给出一系列线段,判断某两个线段是否连通. 思路: 根据线段相交情况建立并查集, 在同一并查集中则连通. (第一反应是强连通分量...实际上只要判断共存即可, 具体的方向啊是没有关系的..) 并查集合并的时候是根节点合并. 快速排斥试验不是必需的, 大规模数据可能是个优化吧. 跨立试验注意共线的情况. 共线判断注意与y 轴平行的情况. #include <cstdio> #include <cstring> #include <cmath> using names…

poj 2653 线段与线段相交

Pick-up sticks Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 11884 Accepted: 4499 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…

POJ 3304 Segments（计算几何：直线与线段相交）

POJ 3304 Segments 大意:给你一些线段,找出一条直线可以穿过全部的线段,相交包含端点. 思路:遍历全部的端点,取两个点形成直线,推断直线是否与全部线段相交,假设存在这种直线,输出Yes.可是注意去重. struct Point { double x, y; } P[210]; struct Line { Point a, b; } L[110]; double xmult(Point p1, Point p2, Point p) { return (p1.x-p.x)*(p2.y…

POJ 3304 Segments (判断直线与线段相交)

题目链接:POJ 3304 Problem 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 at least one point in common. Input…