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.

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

//LCA 模板題，但是我自己想的 bfs+並查集 ，做完後看了看網上的LCA，發現與網上的離線LCA tarjan 算法不一樣，我想的是，遞歸後合並子集合，然後就查找要求的是否在同一個集合，在就輸出即可。

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <vector>
 4 #include <math.h>
 5 #include <string.h>
 6 using namespace std;
 7 #define MX 10005
 8 
 9 int n,m;
10 int a,b;
11 int root,ok;
12 int in[MX]; // 入度
13 vector<int> edge[MX];
14 int f[MX];
15 
16 void Init()
17 {
18     for (int i=1;i<=n;i++)
19         edge[i].clear();
20     memset(in,0,sizeof(in));
21 }
22 
23 int find_head(int x)
24 {
25     return x==f[x]?f[x]:f[x]=find_head(f[x]);
26 }
27 
28 void uni(int x,int y)
29 {
30     x = find_head(x);
31     y = find_head(y);
32     if (x!=y)
33         f[y]=x;     //這個順序很重要，
34 }
35 
36 void tarjan(int x)
37 {
38     for (int i=0;i<(int)edge[x].size();i++)
39     {
40         tarjan(edge[x][i]);
41         uni(x,edge[x][i]);
42         if (ok) return;
43         //每次合並完子集合後，就查詢同一集合
44         if (find_head(a)==find_head(b))
45         {
46             printf("%d\n",x);
47             ok=1;
48             return ;
49         }
50     }
51 }
52 
53 int main()
54 {
55     int T;
56     cin>>T;
57     while (T--)
58     {
59         scanf("%d",&n);
60         Init();
61         m=n-1;
62         for (int i=0;i<m;i++)
63         {
64             int u,v;
65             scanf("%d%d",&u,&v);
66             edge[u].push_back(v);
67             in[v]++;
68         }
69         for (int i=1;i<=n;i++)
70         {
71             if (in[i]==0)
72                 root=i;
73         }
74         scanf("%d%d",&a,&b);
75         for (int i=0;i<=n;i++) f[i]=i;
76         ok=0;
77         tarjan(root);
78     }
79     return 0;
80 }

Nearest Common Ancestors