LCA 学习算法 (最近的共同祖先)poj 1330
Nearest Common Ancestors
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 20983 | Accepted: 11017 |
Description

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
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
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
在求解近期公共祖先为问题上,用到的是Tarjan的思想,从根结点開始形成一棵深搜树。处理技巧就是在回溯到结点u的时候。u的子树已经遍历。这时候才把u结点放入合并集合中,这样u结点和全部u的子树中的结点的近期公共祖先就是u了,u和还未遍历的全部u的兄弟结点及子树中的近期公共祖先就是u的父亲结点。
这样我们在对树深度遍历的时候就非常自然的将树中的结点分成若干的集合,两个集合中的所属不同集合的随意一对顶点的公共祖先都是同样的,也就是说这两个集合的近期公共祖先仅仅有一个。时间复杂度为O(n+q),n为节点,q为询问节点对数。
#include"stdio.h"
#include"string.h"
#include"vector"
using namespace std;
#define N 11000
const int inf=1<<20;
vector<int>g[N];
int s,t,n;
int f[N],pre[N],ans[N];
bool vis[N];
int findset(int x)
{
if(x!=f[x])
f[x]=findset(f[x]);
return f[x];
}
int unionset(int a,int b)
{
int x=findset(a);
int y=findset(b);
if(x==y)
return x;
f[y]=x;
return x;
}
void lca(int u)
{
int i,v;
ans[u]=u;
for(i=0;i<g[u].size();i++)
{
v=g[u][i]; //訪问由父节点引出的各个子节点
lca(v);
int x=unionset(u,v); //把父子节点合并
ans[x]=u; //祖先节点记为U
}
vis[u]=1;
if(u==s&&vis[t]) //两个节点依次被訪问标记,第二次訪问时才满足条件
{
printf("%d\n",ans[findset(t)]);
return ;
}
else if(u==t&&vis[s])
{
printf("%d\n",ans[findset(s)]);
return ;
}
}
int main()
{
int i,u,v,T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(i=1;i<=n;i++)
{
pre[i]=-1; //记录节点I的父节点
f[i]=i; //并查集记录根节点
vis[i]=0;
g[i].clear();
}
for(i=1;i<n;i++)
{
scanf("%d%d",&u,&v);
g[u].push_back(v);
pre[v]=u;
}
scanf("%d%d",&s,&t);
for(i=1;i<=n;i++)
if(pre[i]==-1)
break;
lca(i);
}
return 0;
}
版权声明:本文博客原创文章,博客,未经同意,不得转载。
LCA 学习算法 (最近的共同祖先)poj 1330的更多相关文章
- [最近公共祖先] POJ 1330 Nearest Common Ancestors
Nearest Common Ancestors Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 27316 Accept ...
- 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 ...
- 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 ...
- 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 ...
- POJ 1330 Nearest Common Ancestors (LCA,dfs+ST在线算法)
Nearest Common Ancestors Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 14902 Accept ...
- POJ - 1330 Nearest Common Ancestors(dfs+ST在线算法|LCA倍增法)
1.输入树中的节点数N,输入树中的N-1条边.最后输入2个点,输出它们的最近公共祖先. 2.裸的最近公共祖先. 3. dfs+ST在线算法: /* LCA(POJ 1330) 在线算法 DFS+ST ...
- 最近公共祖先LCA(Tarjan算法)的思考和算法实现
LCA 最近公共祖先 Tarjan(离线)算法的基本思路及其算法实现 小广告:METO CODE 安溪一中信息学在线评测系统(OJ) //由于这是第一篇博客..有点瑕疵...比如我把false写成了f ...
- 最近公共祖先 LCA 倍增算法
树上倍增求LCA LCA指的是最近公共祖先(Least Common Ancestors),如下图所示: 4和5的LCA就是2 那怎么求呢?最粗暴的方法就是先dfs一次,处理出每个点的深度 ...
- POJ 1986 Distance Queries (Tarjan算法求最近公共祖先)
题目链接 Description Farmer John's cows refused to run in his marathon since he chose a path much too lo ...
随机推荐
- 调用ShellExecute需要头文件
调用ShellExecute需要头文件 #include "windows.h " #include "shellapi.h "
- STL__queue_的应用
转:http://hi.baidu.com/xiaotiandm/item/bda34511cf9e99098fbde41a 调用的时候要有头文件: #include<stdlib.h> ...
- Powershell Mail module, 发送outbox 里的全部邮件(一个.csv文件代表一封邮件)
把creating mail代码写到调用处,往outbox写入 mailxxx.csv文件,入面记录了邮件的主要内容 写入 #template $TMP = IMPORT-CSV "$($d ...
- oschina jQuery 插件
jQuery 插件 jQuery自动完成插件(25) jQuery分页插件(20) jQuery 文件上传(21) jQuery 地图插件(14) jQuery对话框(109) jQuery图片展示/ ...
- 自己定义控件(2.2):SurfaceView和SurfaceHolder
本例需求及流程: Activity载入自己定义的SurfaceView-> SurfaceView 构造器中启动线程A.循环改变SurfaceView的x,y坐标,当x,y坐标到某点时设渐显标志 ...
- Andriod中绘(画)图----Canvas的使用具体解释
转载请注明出处:http://blog.csdn.net/qinjuning 因为在网络上找到关于Canvas的使用都比較抽象,或许是我的逻辑思维不太好吧,总是感觉理解起来比較困难, 尤其是对 ...
- 通过Camera进行拍照
Android通过Camera来控制拍照,使用Camera比较简单,按步骤进行即可: 下面用一个示例来演示: Activity: package com.home.activity; import j ...
- 我的Android学习之旅(转)
去年大概在七月份的时候误打误撞接触了一阵子Android,之后由于工作时间比较忙,无暇顾及,九月份的时候自己空闲的时间比较多,公司相对来说加班情况没以前严重.开启了个人的Android学习之旅,初衷是 ...
- [置顶] JQuery实战总结三 标签页效果图实现
在浏览网站时我们会看到当我们鼠标移到多个选项卡上时,不同的选项卡会出现自己对应的界面的要求,在同一个界面上表达了尽量多的信息.大大额提高了空间的利用率.界面的切换效果也是不错的哦,这次自己可以实现啦. ...
- Spring 类构造器初始化实例
构造方法类Bean1 package com.hao947.bean; public class Bean1 { public Bean1() { System.out.println("b ...