Nearest Common Ancestors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 23795   Accepted: 12386 Description A rooted tree is a well-known data structure in computer science and engineering. An example is shown below: In the figure, each…

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++) {  …

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&…

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…

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…

题意:给你一棵树,求两个点的最近公共祖先. 思路:因为只有一组询问,直接数组模拟好了. (写得比较乱) 原题请戳这里 #include <cstdio> #include <bitset> #include <cstring> #include <algorithm> using namespace std; int first[10005],v[10005],next[10005]; int main() { int cas; scanf("%d…

POJ - 1330 Nearest Common Ancestors Time Limit: 1000MS   Memory Limit: 10000KB   64bit IO Format: %lld & %llu Submit Status Description A rooted tree is a well-known data structure in computer science and engineering. An example is shown below:  In t…

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…

POJ.1330 Nearest Common Ancestors (LCA 倍增) 题意分析 给出一棵树,树上有n个点(n-1)条边,n-1个父子的边的关系a-b.接下来给出xy,求出xy的lca节点编号. LCA裸题,用倍增思想. 代码总览 #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> #define nmax 80520 #define demen…

/* *********************************************** Author :kuangbin Created Time :2013-9-5 9:45:17 File Name :F:\2013ACM练习\专题学习\LCA\POJ1330_3.cpp ************************************************ */ #include <stdio.h> #include <string.h> #inclu…