POJ 1330 LCA裸题~

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　POJ 1330　

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
最近在复习一些暑假集训时候学到的一些数据结构，这次的主题是LCA（最近公共祖先），找出一棵树里边任意两个节点的最近公共祖先节点（这个称呼不太科学？），这是一道全裸的LCA的题目，有两种解决思路
方法一：对于每次查询的两个节点，先让两个节点上升到同一个深度的地方，然后两个节点在同时上升，直到两个节点相遇为止，相遇的点即为最近公共祖先。
方法二：先让某个节点一直往上走一直走到根节点，并开一个数组记录这个路径，让后再让另一个节点往上走，直到与前一个节点产生的路径相交为止，那么这个交点也是两个节点的最近公共祖先啦~

方法1的AC代码：

/*********************************
Author: jusonalien
Email : jusonalien@qq.com
school: South China Normal University
Origin:
*********************************/
#include <cstdio>
#include <vector>
#include <cstring>
using namespace std;
const int maxn = 11111;
vector<int>G[maxn];
int depth[maxn],father[maxn];
int root,n;
void dfs(int v,int p,int d){//通过dfs构造出一棵树，并且记录每个节点的深度，这个很重要！
    depth[v] = d;
    for(int i = 0;i < G[v].size();++i){
        if(G[v][i] != p) dfs(G[v][i],v,d+1);
    }
    return ;
}
int lca(int u,int v){
    while(depth[u] > depth[v]) u = father[u];
    while(depth[v] > depth[u]) v = father[v];
    while(u != v){
        u = father[u];
        v = father[v];
    }
    return u;
}
void init(){
    memset(depth,0,sizeof(depth));
    memset(father,-1,sizeof(father));
    for(int i = 1;i <= n;++i) G[i].clear();
}
void print(){//调试代码
    for(int i = 1;i <= n;++i) printf("%02d ",father[i]);
    puts("");
    for(int i = 1;i <= n;++i) printf("%02d ",depth[i]);
    puts("");
}
int main(){
    int cas;
    int a,b;
    scanf("%d",&cas);
    while(cas--){
        scanf("%d",&n);
        init();
        for(int i = 1;i < n;++i){
            scanf("%d%d",&a,&b);
            father[b] = a;
            G[a].push_back(b);
        }
        for(int i = 1;i <= n;++i)
            if(father[i] == -1){
                root = i;break;
            }
        dfs(root,-1,0);
        //print();
        scanf("%d%d",&a,&b);
        printf("%d
",lca(a,b));
    }
    return 0;
}

　方法2的AC代码：

#include <cstdio>
#include <cstring>
using namespace std;
int const maxn = 10000+10;
int fa[maxn];
bool vis[maxn];
int n;
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int u,v;
        scanf("%d",&n);
        memset(vis,0,sizeof(vis));
        memset(fa,0,sizeof(fa));
        for(int i=1;i<n;++i)
        {
            scanf("%d%d",&u,&v);
            fa[v]=u;
        }
        scanf("%d%d",&u,&v);
        do
        {
            vis[u]=true;
            u=fa[u];
        }while(u!=0);
        do
        {
            if(vis[v])
            {
                printf("%d
",v);
                break;
            }
            v=fa[v];
        }while(v!=0);
    }
    return 0;
}

 个人觉得，方法1对于同一棵树上的大规模查询的效率要比方法2要高，并且当查询的节点大都在树的底层的时候，方法2会产生很多不必要的查询，也会产生较多的浪费（并且个人觉得方法1的代码更加优美？

Ps：这里有一份很不错的关于RMQ和LCA的学习资料介绍，请猛戳此处 选自农夫三拳。