LCA入门题集小结

【LCA入门题集小结】的更多相关文章

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2586 题目: How far away ? Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 21500 Accepted Submission(s): 8471 Problem Description There are n house…

poj 1330(RMQ&LCA入门题)

传送门:Problem 1330 https://www.cnblogs.com/violet-acmer/p/9686774.html 参考资料: http://dongxicheng.org/structure/lca-rmq/ 挑战程序设计竞赛(第二版) 变量解释: 对有根树进行DFS,将遍历到的节点按照顺序记下,我们将得到一个长度为2N-1的序列,称之为欧拉序列. total : 记录dfs遍历过程中回溯的节点编号,其实就是从0->2N-1. vs[ ] : 记录DFS访问的顺序,也就是…

POJ1330(LCA入门题)

Nearest Common Ancestors Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 23388 Accepted: 12195 Description A rooted tree is a well-known data structure in computer science and engineering. An example is shown below: In the figure, eac…

lca入门———树上倍增法（博文内含例题）

倍增求LCA: father[i][j]表示节点i往上跳2^j次后的节点可以转移为 father[i][j]=father[father[i][j-1]][j-1] 整体思路: 先比较两个点的深度,如果深度不同,先让深的点往上跳,浅的先不动,等两个点深度一样时,if 相同直接返回,if 不同进行下一步:如果不同,两个点一起跳,j从大到小枚举(其实并不大),如果两个点都跳这么多后,得到的点相等,两个点都不动(因为有可能正好是LCA也有可能在LCA上方),知道得到的点不同,就可以跳上来,然后不…

poj 2524:Ubiquitous Religions（并查集，入门题）

Ubiquitous Religions Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 23997 Accepted: 11807 Description There are so many different religions in the world today that it is difficult to keep track of them all. You are interested in findi…

hrbustoj 1073:病毒（并查集，入门题）

病毒Time Limit: 1000 MS Memory Limit: 65536 KTotal Submit: 719(185 users) Total Accepted: 247(163 users) Rating: Special Judge: NoDescription某种病毒袭击了某地区,该地区有N(1≤N≤50000)人,分别编号为0,1,...,N-1,现在0号已被确诊,所有0的直接朋友和间接朋友都要被隔离.例如:0与1是直接朋友,1与2是直接朋友,则0.2就是间接朋友,那么0.1…