Nearest Common Ancestors POJ - 1330 （LCA）

【Nearest Common Ancestors POJ - 1330 （LCA）】的更多相关文章

Nearest Common Ancestors POJ - 1330 （LCA）

Nearest Common Ancestors Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 34657 Accepted: 17585 Description A rooted tree is a well-known data structure in computer science and engineering. An example is shown below: In the figure, eac…

POJ 1330 （LCA）

http://poj.org/problem?id=1330 题意:给出一个图,求两个点的最近公共祖先. sl :水题,贴个模板试试代码.本来是再敲HDU4757的中间发现要用LCA, 操蛋只好用这个题目试试自己写的对不对. 下面是个倍增的写法,挺实用的. 好了,继续... ; ][MAX],dep[MAX]; G[ G[to].push_back( } dep[u]=d; parent[][u]=pre; ;i<G[u].size();i++) { …

POJ 1330 Nearest Common Ancestors 倍增算法的LCA

POJ 1330 Nearest Common Ancestors 题意:最近公共祖先的裸题思路:LCA和ST我们已经很熟悉了,但是这里的f[i][j]却有相似却又不同的含义.f[i][j]表示i节点的第2j个父亲是多少这个代码不是我的,转自邝斌博客 /* *********************************************** Author :kuangbin Created Time :2013-9-5 9:45:17 File Name :F:\2013AC…

POJ 1330 Nearest Common Ancestors (最近公共祖先LCA + 详解博客）

LCA问题的tarjan解法模板 LCA问题详细 1.二叉搜索树上找两个节点LCA public int query(Node t, Node u, Node v) { int left = u.value; int right = v.value; //二叉查找树内,如果左结点大于右结点,不对,交换 if (left > right) { int temp = left; left = right; right = temp; } while (true) { //如果t小于u.v,往t的右…

POJ 1330 Nearest Common Ancestors (模板题)【LCA】

<题目链接> 题目大意: 给出一棵树,问任意两个点的最近公共祖先的编号. 解题分析:LCA模板题,下面用的是树上倍增求解. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; ; const int INF = 0x3f3f3f3f; struct Edge…

POJ 1330（LCA/倍增法模板）

链接:http://poj.org/problem?id=1330 题意:q次询问求两个点u,v的LCA 思路:LCA模板题,首先找一下树的根,然后dfs预处理求LCA(u,v) AC代码: #include<iostream> #include<algorithm> #include<cmath> #include<cstring> #include<set> #include<string> #include<vector&…

POJ 1330 Nearest Common Ancestors / UVALive 2525 Nearest Common Ancestors （最近公共祖先LCA）

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…