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

Taejon 2002

LCA 模板题
那么在这里讲一下倍增法求LCA好了。
算法分为三步
1. 记录各节点i的深度de[i]。dfs一遍即可。O(N)
2. 预处理出倍增数组，fa[i][j]表示节点i往根的方向跳2^j步的祖先标号。-1表示不存在，也就是跳过根了。fa[i][0]是节点i的父节点标号。那么可以用DP求一下。
3. 对于求点对(u,v)的LCA，如果dep[u]>dep[v]，此时将u往上跳到和v在同一深度。
u,v在同一深度时有两种情况——
1.v本来就是u的祖先，此时u,v相同，lca(u,v) = v；
2.v不是u的祖先，此时u和v距离其lca(u,v)的距离相同，一起往上跳，停下来的位置即是lca(u,v)子节点上。

这就是算法的主要步骤。
这道模板题的代码：

#include <bits/stdc++.h>
#define N 10010
#define M 22
using namespace std;

struct Edge
{
    int next, to, v;
}e[N * 2];
int isroot[N], root;
int n, m, cnt, dep[N], visit[N], fa[N][M], head[N];

inline int read()
{
    int x = 0, f = 1; char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = x * 10 + ch-'0'; ch = getchar(); }
    return x * f;
}

void ins(int u, int v, int w)
{
    e[++ cnt].next = head[u];
    head[u] = cnt;
    e[cnt].to = v;
    e[cnt].v = w;
}

void insert(int u, int v, int w)
{
    ins(u, v, w); 
}

void dfs(int x, int Dep)//第一步
{
    dep[x] = Dep;
    visit[x] = 1;
    for(int i = head[x]; i; i = e[i].next)
        if(visit[ e[i].to ] == 0) dfs(e[i].to, Dep + 1);
}

void First_lca(int n)//第二步
{
    for(int j = 1; j < M; j ++)
        for(int i = 1; i <= n; i ++)
            if(~fa[i][j - 1])
                fa[i][j] = fa[ fa[i][j - 1] ][j - 1];
}

int Find_lca(int x, int y)//第三步
{
    if(dep[x] < dep[y]) swap(x, y);
    for(int b = M - 1; b >= 0; b --)
        if(dep[ fa[x][b] ] >= dep[y])
            x = fa[x][b];
    if(x == y) return x;
    for(int b = M - 1; b >= 0; b --)
        if(fa[x][b] != fa[y][b])
            x = fa[x][b], y = fa[y][b];
    return fa[x][0];
}

int main()
{
    int T;
    T = read();
    while(T --)
    {
        memset(head, 0, sizeof(head));
        memset(isroot, 0, sizeof(isroot));
        memset(visit, 0, sizeof(visit));
        memset(fa, 0xff, sizeof(fa));
        memset(dep, 0, sizeof(dep));
        cnt = 0;
        n = read();
        int x, y;
        for(int i = 1; i < n; i ++)
        {
            x = read();
            y = read();
            isroot[y] = 1;
            insert(x, y, 0);
            fa[y][0] = x;
        }
        for(int i = 1; i <= n; i ++)
            if(isroot[i] == 0) {root = i;break;}
        dfs(root, 1);
        First_lca(n);
        x = read();
        y = read();
        printf("%d\n", Find_lca(x, y));
    }
    return 0;
}