[置顶] Nearest Common Ancestors（POJ-1330）(LCA转RMQ在线算法)

Nearest Common Ancestors

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 29996		Accepted: 15332

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

题目链接：http://poj.org/problem?id=1330

题目大意： 这个题目的意思就是读入有n个顶点，n-1条边的树，在给出一个查询，求得这两个顶点的的最小公共祖先。
   最小公共祖
c470
先就是距离两个点最近的父节点。
题目思路： 最基础的LCA模板题只有一个查询，最近学了LCA转RMQ在线算法就直接写了，虽然错了好多次。。
                   但是这个最容易想到的就是直接从底层向上找到同一个父节点为止，这个写了也给贴出来啦。

最容易想到的写法：

代码1：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <vector>
using namespace std;
typedef long long LL;
const int N=10099;
int dpth
; //存储每个节点的深度
int per
; // 存储每个节点的祖先
vector <int> a
;
void dfs(int x,int d)
{
dpth[x]=d;
for(int i=0; i<a[x].size(); i++)
{
dfs(a[x][i],d+1);
}
}
int main()
{
int n,t;
scanf("%d",&t);
while(t--)
{
memset(per,-1,sizeof(per)); //将祖先初始化为-1
scanf("%d",&n);
for(int i=1; i<=n; i++) a[i].clear(); //将vector数组清空
int x,y;
for(int i=1; i<=n-1; i++)
{
scanf("%d%d",&x,&y);
a[x].push_back(y); //将x->y存入vector数组
per[y]=x;
}
int s=1;
while(per[s]>=1) //找到树的根节点
s++;
dfs(s,0);
scanf("%d%d",&x,&y);
//一层一层的向上找，直到找到共同祖先跳出
while(x!=y)
{
if(dpth[x]>dpth[y]) x=per[x];
else y=per[y];
}
printf("%d\n",x);
}
return 0;
}

LCA写法 ：我觉得比较麻烦，要建立好多数组。。
贴个看着不错的博客的链接，不懂LCA的可以去看看。
推荐一下：LCA在线算法ST算法

代码2：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <vector>
using namespace std;
typedef long long LL;
const int N=10099;
int dp[N<<1][30],dpth[N<<1];    // dp数组用RMQ计算深度的最小值下标 ，dpth数组存储深度
int first
,vis[N<<1],k,per
;//first数组标记节点第一次出现，vis数组存储遍历的顺序，per数组存储祖先
bool book
;                   //book数组用来标记这个点进入过没有
vector <int> a
;
//最小深度的下标
void ST(int n)
{
for(int i=1;i<=n;i++)
dp[i][0]=i;
for(int j=1;(1<<j)<=n;j++)
{
for(int i=1 ; i+(1<<j)-1<=n ; i++)
{
int a=dp[i][j-1];
int b=dp[i+(1<<(j-1))][j-1];
if(dpth[a]<=dpth[b]) dp[i][j]=a;
else dp[i][j]=b;
}
}
}
int RMQ(int l,int r)
{
int k=log(r-l+1.0)/log(2.0);
int a=dp[l][k];
int b=dp[r-(1<<k)+1][k];
if(dpth[a]<=dpth[b]) return a;
else return b;
}
int LCA(int u,int v)
{
//找出这两个数在first数组中的值，排列成 x<y
int x=first[u];
int y=first[v];
if(x>y) swap(x,y);
int res=RMQ(x,y);
return vis[res]; //返回在遍历数组中对应的节点
}
void dfs(int x,int d)
{
book[x]=true;
k++;
first[x]=k;
dpth[k]=d;
vis[k]=x;
for(int i=0; i < a[x].size() ;i++)
{
if(!book[a[x][i]])
{
dfs(a[x][i],d+1);
k++;
dpth[k]=d;
vis[k]=x;
}
}
}
int main()
{
int n,t;
scanf("%d",&t);
while(t--)
{
//初始化
memset(book,false,sizeof(book));
memset(vis,0,sizeof(vis));
memset(per,-1,sizeof(per));
k=0;
scanf("%d",&n);
for(int i=1; i<=n; i++) a[i].clear();
int x,y;
for(int i=1; i<=n-1; i++)
{
scanf("%d%d",&x,&y);
a[x].push_back(y);
per[y]=x;
}
int i;
for(i=1;i<=n-1;i++) if(per[i]==-1) break; //找根节点
dfs(i,0);
ST(2*n-1); //遍历n个节点刚好2n-1
scanf("%d%d",&x,&y);
int ans=LCA(x,y);
printf("%d\n",ans);
}
return 0;
}