题目链接:http://poj.org/problem?id=1330

Nearest Common Ancestors
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 36918   Accepted: 18495

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

Source

 
题目大意:输入T  T组样例  输入N  N个结点(1-N) 下面N-1行  每行两个数 u v  表示u是v的父亲  第N行表示询问 两个数的最近公共祖先
思路:不多说,完全板子
看代码:
#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
using namespace std;
const int maxn=1e4+;
int N;//节点个数
vector<int>v[maxn];//树
vector<int> query[maxn];
int indeg[maxn];//节点的入度
int fa[maxn],deep[maxn],ancestor[maxn];//父亲 深度 祖先
bool vis[maxn];//是否被检查过
int root;
void Init()
{
for(int i=;i<=N;i++)
{
v[i].clear();
query[i].clear();
indeg[i]=; }
return ;
}
void add_edge(int x,int y)
{
v[x].push_back(y);
indeg[y]++;
return ;
}
void Input_query()
{
int u,v;
scanf("%d%d",&u,&v);
query[u].push_back(v);//注意 两个都要存
query[v].push_back(u);
return ;
}
void Init_set()
{
for(int i=;i<=N;i++)
{
fa[i]=i;
ancestor[i]=i;
deep[i]=;
}
return ;
}
int Find(int x)
{
return fa[x]==x?x:fa[x]=Find(fa[x]);
}
void Union(int u,int v)
{
int du=Find(u);
int dv=Find(v);
if(du>dv)
{
fa[dv]=du;
return ;
}
else
{
fa[du]=dv;
if(deep[du]==deep[dv]) deep[dv]++;
}
return ;
}
void Tarjan(int p)
{
for(int i=;i<v[p].size();i++)//遍历子树
{
Tarjan(v[p][i]);
Union(p,v[p][i]);
ancestor[Find(p)]=p;
}
vis[p]=true;
for(int i=;i<query[p].size();i++)
{
if(vis[query[p][i]])
{
printf("%d\n",ancestor[Find(query[p][i])]);
}
}
return ;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{ scanf("%d",&N);
Init();
for(int i=;i<N;i++)
{
int u,v;
scanf("%d%d",&u,&v);
add_edge(u,v);
}
for(int i=;i<=N;i++)
{
if(indeg[i]==)
{
root=i;
break;
}
}
Input_query();
Init_set();
memset(vis,false,sizeof(vis));
Tarjan(root);
}
return ;
}

Nearest Common Ancestors(LCA板子)的更多相关文章

  1. 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 ...

  2. POJ 1330 Nearest Common Ancestors LCA题解

    Nearest Common Ancestors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 19728   Accept ...

  3. pku 1330 Nearest Common Ancestors LCA离线

    pku 1330 Nearest Common Ancestors 题目链接: http://poj.org/problem?id=1330 题目大意: 给定一棵树的边关系,注意是有向边,因为这个WA ...

  4. poj 1330 Nearest Common Ancestors lca 在线rmq

    Nearest Common Ancestors Description A rooted tree is a well-known data structure in computer scienc ...

  5. poj 1330 Nearest Common Ancestors LCA

    题目链接:http://poj.org/problem?id=1330 A rooted tree is a well-known data structure in computer science ...

  6. Nearest Common Ancestors(LCA)

    Description A rooted tree is a well-known data structure in computer science and engineering. An exa ...

  7. [POJ1330]Nearest Common Ancestors(LCA, 离线tarjan)

    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 ...

  8. POJ 1330 Nearest Common Ancestors(LCA模板)

    给定一棵树求任意两个节点的公共祖先 tarjan离线求LCA思想是,先把所有的查询保存起来,然后dfs一遍树的时候在判断.如果当前节点是要求的两个节点当中的一个,那么再判断另外一个是否已经访问过,如果 ...

  9. POJ 1330 Nearest Common Ancestors (LCA,倍增算法,在线算法)

    /* *********************************************** Author :kuangbin Created Time :2013-9-5 9:45:17 F ...

随机推荐

  1. unittest测试框架详谈及实操(五)

    测试报告——生成HTML格式的测试报告 前面的实例输出的所有测试结果都是以命令行日志的方式展示,不止于难看,但也不适合直接把那样的测试结果截图发给相关人员,尤其是领导.这时需要更加友好的测试结果,既能 ...

  2. Java内存模型(转载)

    本文章节: 1.JMM简介 2.堆和栈 3.本机内存 4.防止内存泄漏 1.JMM简介 i.内存模型概述 Java平台自动集成了线程以及多处理器技术,这种集成程度比Java以前诞生的计算机语言要厉害很 ...

  3. Jenkins构建完成之后运行脚本可以杀掉TomCat但是起不来的解决方法

    Jenkins构建完成之后运行脚本可以杀掉TomCat但是起不来的解决方法 写了一个重启tomcat的脚本,让jenkins编译.打包.发布时调用.在本地写好重启tomcat的脚本后,本地执行脚本没有 ...

  4. C#中使用Redis学习一 windows安装redis服务器端和客户端

    学习背景 今天是2015年1月2日,新年刚开始的第二天,先祝大家元旦快乐啦(迟到的祝福吧^_^).前段时间一直写Jquery插件开发系列博文,这个系列文章暂停一段时间,最近一直在看redis,我将把r ...

  5. asp.net——地址栏传递中文参数乱码解决方案

    地址栏传递中文参数乱码解决方案: 很多人在使用地址栏传递参数的时候都会遇到一个麻烦的问题(参数为中文时乱码了),那要怎么解决呢? 其实解决这个问题也不怎么难,无非就是给要传递的中文参数一个编码解码的过 ...

  6. Windows上编译libpng

    下载libpng 1.5.10并解压到[工作目录]/png/libpng-1.5.10 用CMake选择png/libpng-1.5.10目录并Configure: CMAKE_C_FLAGS_DEB ...

  7. webapi发布常见错误及解决方案

    webapi发布常见错误及解决方案 错误一: 错误:404 (Not Found) 解决方案: 在  <system.webServer>节点中添加如下模块: <modules ru ...

  8. 「HNOI 2013」游走

    题目链接 戳我 \(Solution\) 首先申明几个变量: f[x]:到点x的概率, vis[x]:x点的度 dp[x][y]:(x,y)这条边的概率 number[x][y]:x这条边的编号 下面 ...

  9. 【洛谷九月月赛T1】签到题(bsgs)(快速乘)

    说好的签到题呢qwq....怎么我签到题都不会啊qwq 之后看了bsgs才发现貌似不是那么那么难fake!!什么东西... 先贴上部分分做法(也就是枚举1的个数,然后每一步都进行取模(这和最后取模结果 ...

  10. Django-05模型层之单表操作1

    7.1 ORM简介 MVC或者MVC框架中包括一个重要的部分,就是ORM,它实现了数据模型与数据库的解耦,即数据模型的设计不需要依赖于特定的数据库,通过简单的配置就可以轻松更换数据库,这极大的减轻了开 ...