描述 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 connecte…

转载请注明出处: http://www.cnblogs.com/fraud/          ——by fraud Jack Straws Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 3512   Accepted: 1601 Description In the game of Jack Straws, a number of plastic or wooden "straws" are dumped o…

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…

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 <…

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…

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…

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…

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:要求相交的线段都要塞进同一个集合里 sol:并查集+判断线段相交即可.n很小所以n^2就可以水过 #include <iostream> #include <cmath> #include <cstring> #include <cstdio> using namespace std; ]; char ch; int tmp,n; double X1,X2,Y1,Y2; #define eps 1e-8 #define PI acos(-1.0)//3…

先说一下题目大意:给定一些线段,这些线段顺序编号,这时候如果两条线段相交,则把他们加入到一个集合中,问给定一个线段序号,求在此集合中有多少条线段. 这个题的难度在于怎么判断线段相交,判断玩相交之后就是怎么找个他们之间的联系,这时候就要用到并查集了. 步骤: 1.判断两条线段相交 2. 用并查集实现查找线段个数和添加到集合中 关于这个判断线段相交的问题.我搞了一晚上加上一下午,刚开始自己想了一种数学上的相交,就是先求出两条线段所在的线性方程,然后求出他们的交点,最后在判断这个交点在不在这两个线段之…