概念: 对于有根树T的两个节点u,v,最近公共祖先LCA(T, u, v)表示一个节点 x, 满足 x 是 u , v 的祖先且 x 的深度尽可能的大.即从 u 到 v 的路径一定经过点 x. 算法: 解决LCA问题比较经典的是Tarjan - LCA 离线算法,还有另外一种方法,是经过一系列处理将LCA问题转化为和数据结构有关的RMQ问题加以解决.这里只阐述下Tarjan - LCA 算法. Tarjan - LCA算法: 此算法基于 DFS 框架,每搜到一个新的节点,就创建由这个节点构成的集…
#include"stdio.h" #include"string.h" #include"iostream" #include"queue" #define M 111111 using namespace std; struct st { int u,v,next,w; }edge[M*2]; int rank[M],head[M],t,pre[M],use[M],dis[M]; void init() { t=0; me…
POJ 1470 Closest Common Ancestors(最近公共祖先 LCA) Description Write a program that takes as input a rooted tree and a list of pairs of vertices. For each pair (u,v) the program determines the closest common ancestor of u and v in the tree. The closest co…
POJ 1330 Nearest Common Ancestors / UVALive 2525 Nearest Common Ancestors (最近公共祖先LCA) Description A rooted tree is a well-known data structure in computer science and engineering. An example is shown below: In the figure, each node is labeled with an…
洛谷 SP14932 LCA - Lowest Common Ancestor 洛谷评测传送门 题目描述 A tree is an undirected graph in which any two vertices are connected by exactly one simple path. In other words, any connected graph without cycles is a tree. - Wikipedia The lowest common ancesto…
高级的算法——倍增!!! 根据LCA的定义,我们可以知道假如有两个节点x和y,则LCA(x,y)是 x 到根的路 径与 y 到根的路径的交汇点,同时也是 x 和 y 之间所有路径中深度最小的节 点,所以,我们可以用遍历路径的方法求 LCA. 但想想都知道啦,这种遍历的方法肯定too slow,最坏情况时可达到O(n),数据大点儿,就光荣TLE了. 所以我们高级的化身——倍增算法就出现了! 谈谈倍增—— 倍增简单来讲就是两个点跳到同一高度后,再一起往上跳,直到跳到一个共同的点,就能找到它们的最近公…
Nearest Common Ancestors Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 20715 Accepted: 10910 Description A rooted tree is a well-known data structure in computer science and engineering. An example is shown below: In the figure, each…
转载来自:https://blog.andrewei.info/2015/10/08/e6-9c-80-e8-bf-91-e5-85-ac-e5-85-b1-e7-a5-96-e5-85-88lca-e7-9a-84-e4-b8-89-e7-a7-8d-e6-b1-82-e8-a7-a3-e6-96-b9-e6-b3-95/ 简述 LCA(Least Common Ancestors),即最近公共祖先,是指这样一个问题:在有根树中,找出某两个结点 u 和 v 最近的公共祖先(另一种说法,离树根最…
A rooted tree is a well-known data structure in computer science and engineering. An example is shown below: In the figure, each node is labeled with an integer from {1, 2,...,16}. Node 8 is the root of the tree. Node x is an ancestor of node y if n…