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. 数学公式在 iOS 中的表示

    1. 三角函数  double sin (double);正弦  double cos (double);余弦  double tan (double);正切 2 .反三角函数  double asi ...

  2. ORACLE 判断首字母大小写问题

    1.对判断的字段进行拆分 select  substr(要区分的字段,0,1)  from 表 : 得到一个 首字母 2.对这个字符进行大小写判断 查出以小写字符为开头的 select  substr ...

  3. Linux centosVMware 负载均衡集群介绍、LVS介绍、LVS调度算法、LVS NAT模式搭建

    一.负载均衡集群介绍 主流开源软件LVS.keepalived.haproxy.nginx等 其中LVS属于4层(网络OSI 7层模型),nginx属于7层,haproxy既可以认为是4层,也可以当做 ...

  4. Linux centosVMware NFS exportfs命令、NFS客户端问题、FTP介绍、使用vsftpd搭建ftp

    一.exportfs命令 常用选项 -a 全部挂载或者全部卸载 -r 重新挂载 -u 卸载某一个目录 -v 显示共享目录 以下操作在服务端上 vim /etc/exports //增加 /tmp/ 1 ...

  5. Linux centosVMware Nginx负载均衡、ssl原理、生成ssl密钥对、Nginx配置ssl

    一.Nginx负载均衡 vim /usr/local/nginx/conf/vhost/load.conf // 写入如下内容 upstream qq_com { ip_hash; 同一个用户始终保持 ...

  6. redhat 7.6 密码破解(无光盘)

    开机,在下面界面按e 找到linux16  在最尾输入 rd.break 按 Ctrl+x 输入 mount -o remount,rw /sysroot 输入chroot   /sysroot sh ...

  7. vue中配置sass(包含vue-cli 3)

    目录 vue vue cli 3 老版本的脚手架搭建的项目 版本 安装 不用修改任何配置 vue文件中使用 vue 更新时间: 2018-09-21 vue cli 3 选择 Manually sel ...

  8. titleView发生偏移、titleView与masonry、titleView的设置、titleView的使用

    navigationItem的titleView属性的设置本身是很简单的,容易出问题的原因是自动化布局与frame混用造成的. 本文一步一步的讲解,力求找到问题的起源.如果你也在这块同样遇到问题,不妨 ...

  9. c++中的全排列

    next_permutation函数 组合数学中经常用到排列,这里介绍一个计算序列全排列的函数:next_permutation(start,end),和prev_permutation(start, ...

  10. Day3-N - Monthly Expense POJ3273

    Farmer John is an astounding accounting wizard and has realized he might run out of money to run the ...