1143 Lowest Common Ancestor (30 分)

本文介绍了一种算法,用于在给定的二叉搜索树中找到任意两个节点的最低公共祖先(LCA)。通过分析树的性质和节点之间的关系,该算法能够有效地解决这一问题,并提供了输入输出规范以及一个示例输入输出。

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The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U and V as descendants.

A binary search tree (BST) is recursively defined as a binary tree which has the following properties:

  • The left subtree of a node contains only nodes with keys less than the node's key.
  • The right subtree of a node contains only nodes with keys greater than or equal to the node's key.
  • Both the left and right subtrees must also be binary search trees.

Given any two nodes in a BST, you are supposed to find their LCA.

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers: M (≤ 1,000), the number of pairs of nodes to be tested; and N (≤ 10,000), the number of keys in the BST, respectively. In the second line, N distinct integers are given as the preorder traversal sequence of the BST. Then M lines follow, each contains a pair of integer keys U and V. All the keys are in the range of int.

Output Specification:

For each given pair of U and V, print in a line LCA of U and V is A. if the LCA is found and A is the key. But if A is one of U and V, print X is an ancestor of Y. where X is A and Y is the other node. If U or V is not found in the BST, print in a line ERROR: U is not found. or ERROR: V is not found. or ERROR: U and V are not found..

Sample Input:

6 8
6 3 1 2 5 4 8 7
2 5
8 7
1 9
12 -3
0 8
99 99

Sample Output:

LCA of 2 and 5 is 3.
8 is an ancestor of 7.
ERROR: 9 is not found.
ERROR: 12 and -3 are not found.
ERROR: 0 is not found.
ERROR: 99 and 99 are not found.

 #include<bits/stdc++.h>
using namespace std;
map<int,bool>a;
int main()
{
    int m,n,u,v,x;
    cin>>m>>n;
    vector<int>t(n);
    for(int i=0;i<n;i++)
    {
        cin>>t[i];
        a[t[i]]=true;
    }
    for(int i=0;i<m;i++)
    {
        cin>>u>>v;
        for(int j=0;j<n;j++)
        {
            x=t[j];
            if((x>=u&&x<=v)||(x>=v&&x<=u))
                break;
        }
        if(a[u]==false&&a[v]==false)
        {
            printf("ERROR: %d and %d are not found.\n",u,v);
        }
        else if(a[u]==false||a[v]==false)
        {
            if(a[u]==false)
                printf("ERROR: %d is not found.\n",u);
            else
                printf("ERROR: %d is not found.\n",v);
        }
        else if(x==u||x==v)
        {
            if(x==u)
                printf("%d is an ancestor of %d.\n",u,v);
            else
                printf("%d is an ancestor of %d.\n",v,u);
        }
        else
            printf("LCA of %d and %d is %d.\n",u,v,x);
    }
    return 0;
}

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