Nearest Common Ancestors
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 20940   Accepted: 11000

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 integer from {1, 2,...,16}. Node 8 is the root of the tree. Node x is an ancestor of node y if node x is in the path between the root and node y. For example, node 4 is an ancestor of node 16. Node 10 is also an ancestor
of node 16. As a matter of fact, nodes 8, 4, 10, and 16 are the ancestors of node 16. Remember that a node is an ancestor of itself. Nodes 8, 4, 6, and 7 are the ancestors of node 7. A node x is called a common ancestor of two different nodes y and z if node
x is an ancestor of node y and an ancestor of node z. Thus, nodes 8 and 4 are the common ancestors of nodes 16 and 7. A node x is called the nearest common ancestor of nodes y and z if x is a common ancestor of y and z and nearest to y and z among their common
ancestors. Hence, the nearest common ancestor of nodes 16 and 7 is node 4. Node 4 is nearer to nodes 16 and 7 than node 8 is. 



For other examples, the nearest common ancestor of nodes 2 and 3 is node 10, the nearest common ancestor of nodes 6 and 13 is node 8, and the nearest common ancestor of nodes 4 and 12 is node 4. In the last example, if y is an ancestor of z, then the nearest
common ancestor of y and z is y. 



Write a program that finds the nearest common ancestor of two distinct nodes in a tree. 


Input

The input consists of T test cases. The number of test cases (T) is given in the first line of the input file. Each test case starts with a line containing an integer N , the number of nodes in a tree, 2<=N<=10,000. The nodes are labeled with integers 1, 2,...,
N. Each of the next N -1 lines contains a pair of integers that represent an edge --the first integer is the parent node of the second integer. Note that a tree with N nodes has exactly N - 1 edges. The last line of each test case contains two distinct integers
whose nearest common ancestor is to be computed.

Output

Print exactly one line for each test case. The line should contain the integer that is the nearest common ancestor.

Sample Input

2
16
1 14
8 5
10 16
5 9
4 6
8 4
4 10
1 13
6 15
10 11
6 7
10 2
16 3
8 1
16 12
16 7
5
2 3
3 4
3 1
1 5
3 5

Sample Output

4
3

并查集。之前在hihoCoder第十二周做过类似的。

代码:

#include <iostream>
#include <vector>
#include <string>
#include <cstring>
#include <algorithm>
#pragma warning(disable:4996)
using namespace std; int father[10005]; void result(int test1,int test2)
{
int node2=test2;
while(father[test1]!=test1)
{
node2=test2;
while(father[node2]!=node2)
{
if(test1==node2)
{
cout<<test1<<endl;
return;
}
node2=father[node2];
}
test1=father[test1];
}
cout<<test1<<endl;
return;
} int main()
{
int count;
cin>>count; while(count--)
{
int fa,son,node,i;
cin>>node; for(i=1;i<10005;i++)
{
father[i]=i;
} for(i=1;i<=node-1;i++)
{
cin>>fa>>son;
father[son]=fa;
}
int test1,test2;
cin>>test1>>test2; result(test1,test2);
}
return 0;
}

版权声明:本文为博主原创文章,未经博主允许不得转载。

POJ 1330:Nearest Common Ancestors的更多相关文章

  1. 【51.64%】【POJ 1330】Nearest Common Ancestors

    Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 26416 Accepted: 13641 Description A roote ...

  2. 【Poj 1330】Nearest Common Ancestors

    http://poj.org/problem?id=1330 题目意思就是T组树求两点LCA. 这个可以离线DFS(Tarjan)-----具体参考 O(Tn) 0ms 还有其他在线O(Tnlogn) ...

  3. 【POJ 1330】 Nearest Common Ancestors

    [题目链接] 点击打开链接 [算法] 倍增法求最近公共祖先 [代码] #include <algorithm> #include <bitset> #include <c ...

  4. POJ 1330 Nearest Common Ancestors(Tree)

    题目:Nearest Common Ancestors 根据输入建立树,然后求2个结点的最近共同祖先. 注意几点: (1)记录每个结点的父亲,比较层级时要用: (2)记录层级: (3)记录每个结点的孩 ...

  5. POJ 1330 Nearest Common Ancestors 【LCA模板题】

    任意门:http://poj.org/problem?id=1330 Nearest Common Ancestors Time Limit: 1000MS   Memory Limit: 10000 ...

  6. 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 ...

  7. POJ - 1330 Nearest Common Ancestors(基础LCA)

    POJ - 1330 Nearest Common Ancestors Time Limit: 1000MS   Memory Limit: 10000KB   64bit IO Format: %l ...

  8. 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 ...

  9. POJ.1330 Nearest Common Ancestors (LCA 倍增)

    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 ...

随机推荐

  1. onContextItemSelected 与 onMenuItemSelected 的那些事

    Android 的activity中onCreateOptionsMenu onMenuItemSelected onOptionsItemSelected onCreateContextMenu o ...

  2. Nodejs回调加超时限制两种实现方法

    odejs回调加超时限制两种实现方法 Nodejs下的IO操作都是异步的,有时候异步请求返回太慢,不想无限等待回调怎么办呢?我们可以给回调函数加一个超时限制,到一定时间还没有回调就表示失败,继续后面的 ...

  3. H.264 中的Annex B格式和AVCC格式

    首先要理解的是没有标准的H.264基本流格式.文档中的确包含了一个Annex,特别是描述了一种可能的格式Annex B格式,但是这个并不是一个必须要求的格式.标准文档中指定了视频怎样编码成独立的包,但 ...

  4. ABC155E - Payment

    简述题意,给你一个大数,你可以选择10的次幂进行加减运算,问如何用最少的次数从0到达这个大数 考虑从这个大数到0,从最低位开始,每次都将这个位置取完,2种策略,贪心的话不好处理进位的情况,可以想到是D ...

  5. 「CH6801」棋盘覆盖

    「CH6801」棋盘覆盖 传送门 考虑将棋盘黑白染色,两个都无障碍的相邻的点之间连边,边的容量都为1,然后就求一次最大匹配即可 参考代码: #include <cstring> #incl ...

  6. Python 之网络编程之进程总体概要

     一: 进程的概念:(Process) 进程就是正在运行的程序,它是操作系统中,资源分配的最小单位. 资源分配:分配的是cpu和内存等物理资源 进程号是进程的唯一标识 同一个程序执行两次之后是两个进程 ...

  7. Hive的存储和MapReduce处理——数据清洗(Part3)

    日期:2019.11.17 博客期:118 星期日 这几天在写程序的时候虚拟机崩了,无语~所以重新从最初的状态开始配环境,重新整理之前的所有代码程序.

  8. ecshop代码分析一(init.php文件)

    ecshop代码分析一(init.php文件)   因为工作原因,需要对ecshop二次开发,顺便记录一下对ecshop源代码的一些分析: 首先是init.php文件,这个文件在ecshop每个页面都 ...

  9. vue cli3.0打包

    1.vue cli3.0需要在项目根目录下配置webpack  包括反向代理以及打包文件路径 const webpack = require("webpack"); module. ...

  10. 关于Redis 分布式 微服务 集群Cluster

    一:Redis 1,redis是一个高性能的键值对存储方式的数据库,同时还提供list,set,zset,hash等数据结构的存储. 2,Redis运行在内存中但是可以持久化到磁盘,所以在对不同数据集 ...