A - 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 the figu…

Nearest Common Ancestors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 18136   Accepted: 9608 Description A rooted tree is a well-known data structure in computer science and engineering. An example is shown below: In the figure, each…

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…

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…

  Nearest Common Ancestors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 24587   Accepted: 12779 Description A rooted tree is a well-known data structure in computer science and engineering. An example is shown below:  In the figure, e…

Nearest Common Ancestors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 37386   Accepted: 18694 Description A rooted tree is a well-known data structure in computer science and engineering. An example is shown below: In the figure, each…

题目链接:http://poj.org/problem?id=1330 题意就是求一组最近公共祖先,昨晚学了离线tarjan,今天来实现一下. 个人感觉tarjan算法是利用了dfs序和节点深度的关系,大致的意思:dfs如果不递归到递归基,那么dfs就会越递归越深,这个时候深度也是相应增加的,所以这个时候任意在已经遍历过的节点中选取两个点,计算他们的lca也就相当于是用并查集求他们的root.而dfs执行到递归基,转而执行下一个分支的时候,这个时候dfs的节点应当是小于等于之前执行到递归基的节点…

http://poj.org/problem? id=1330 给一个有根树,一个查询节点(u,v)的近期公共祖先 836K 16MS #include<iostream> #include<cstring> #include<cstdio> #include<algorithm> #include<vector> #include<string> #include<set> #include<map> con…

经典LCA操作.. 贴AC代码 import java.lang.reflect.Array; import java.util.*; public class POJ1330 { // 并查集部分 static int[] p; static int find(int u){ if(p[u]!=u){ p[u] = find(p[u]); } return p[u]; } static class Edge{ int to; int next; } // 链式前向星 static int[]…

该算法的详细解释请戳: http://www.cnblogs.com/Findxiaoxun/p/3428516.html #include<cstdio> #include<algorithm> #include<cstring> #include<vector> using namespace std; ; int father[MAXN],ancestor[MAXN]; bool visit[MAXN]; int ans[MAXN]; vector&l…