An Easy Problem?! Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11576   Accepted: 1760 Description It's raining outside. Farmer Johnson's bull Ben wants some rain to water his flowers. Ben nails two wooden boards on the wall of his bar…

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…

题意:给定n个木棍依次放下,要求最终判断没被覆盖的木棍是哪些. 思路:快速排斥以及跨立实验可以判断线段相交. #include<algorithm> #include<cstdio> #include<cmath> #include<cstring> #include<iostream> ; struct Point{ double x,y; Point(){} Point(double x0,double y0):x(x0),y(y0){} }…

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…

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…

Pick-up sticks Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 12861   Accepted: 4847 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…

Treasure Hunt Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 7857   Accepted: 3247 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…

题意:给定一堆线段,然后有询问,问这两个线段是不是相交,并且如果间接相交也可以. 析:可以用并查集和线段相交来做,也可以用Floyd来做,相交就是一个模板题. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #includ…

题意: 有n个木棍,给出木棍的两个端点的x,y坐标,判断其中某两个线段是否连通(可通过其他线段连通) #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <vector> #include <queue> #include <map> #include <…

题意: 题意很好理解,从左边射过来的光线,最远能经过管道到右边多少距离. 分析: 光线一定经过一个上端点和一个下端点,这一点很容易想到.然后枚举上下端点即可 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #define eps 1e-8 #define INF 1e9 #define OK sgn(tmp…