POJ 1330 Nearest Common Ancestors "Recent public ancestor LCA algorithm +tarjan offline algorithm"

Nearest Common Ancestors

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 20715		Accepted: 10910

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 a labeled with a integer from {1, 2,..., 16}. Node 8 is the root of the tree. Node x is a ancestor of node y if node x is in the path between the root and node Y. For example, node 4 was an ancestor of node 16. Node also an ancestor of node 16. As a matter of fact, nodes 8, 4, 16, and + are the ancestors of the node. Remember that a node was 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 the 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 and 7. A node x is called the nearest common ancestor of nodes Y and Z if x are a common ancestor of Y and Z and nearest to Y and Z among their common ancestors. Hence, the nearest common ancestor of nodes and 7 is node 4. Node 4 is nearer-nodes and 7 than node 8 is.

For other examples, the nearest common ancestor of nodes 2 and 3 are node, the nearest common ancestor of nodes 6 and 13 is node 8, and the nearest common ancestor of nodes 4 and are node 4. The last example, if Y is a ancestor of Z, then the nearest common ancestor of Y and Z are y.

Write A program This finds the nearest common ancestor of the distinct nodes in a tree.

Input

The input consists of T test cases. The number of test cases (T) is given on the first line of the input file. Each test case is starts with a line containing an integer N and the number of nodes in a tree, 2<=n<=10,000. The nodes is labeled with integers 1, 2,..., N. Each of the next N-1 lines contains a pair of integers this represent an edge--the first integer is the parent node of T He second integer. Note that a tree with N nodes have exactly N-1 edges. The last line of all test case contains, distinct integers whose nearest common ancestor are to be computed.

Output

Print exactly one line for each test case. The line should contain the "the integer" is the nearest common ancestor.

Sample Input

2161 148 510 165 94 68 44 101 136 1510 116 710 216 38 116 1216 752 33 43 11 53 5

Sample Output

43 topic Analysis: T Group data, each group has n nodes, N-1 edge, so it must be a tree. The last line of input for each group is two points U, v. Who is the nearest public ancestor to you and V? 
 Tanjan offline algorithm. 
  

#include <stdio.h> #include <string.h> #include <stdlib.h> #include <math.h> #include < vector> #include <algorithm> #define N 10000+10using namespace Std;int n; int s, e;vector<int>q[n];int fa[n];bool Vis[n];bool root[n];//mark the point is not the root node int findset (int x)//compressed path and check set {return fa[x ]!=x?fa[x]=findset (Fa[x]): x;}        void LCA (int u) {for (int i=0; i<q[u].size (); i++) {LCA (q[u][i]);    if (findset (u) = Findset (Q[u][i])) {Fa[fa[q[u][i]]] = fa[u];//merge}} Vis[u]=true;        if (u==s && vis[e]==true) {printf ("%d\n", Findset (e));    return;        } if (U==e && vis[s]==true) {printf ("%d\n", Findset (s));    return;    }}int Main () {int t;    scanf ("%d", &t);    int I, j, K;    int u, v;            while (t--) {scanf ("%d", &n),//n node//initialize for (i=0; i<=n; i++) {q[i].clear (); Fa[i]=i;    Set the Father node as its own root[i]=true;        Vis[i]=false;            Tag not accessed} for (i=0; i<n-1; i++) {scanf ("%d%d", &u, &v);//u is the Father node of V            Q[u].push_back (v);        Root[v]=false;        } scanf ("%d%d", &s, &e);                for (I=1; i<=n; i++) {if (root[i]==true)//The point is the root node {LCA (i);//LCA one off-line algorithm            Break }}}return 0;}

POJ 1330 Nearest Common Ancestors "Recent public ancestor LCA algorithm +tarjan offline algorithm"