1143 Lowest Common Ancestor （30 分）（C++）

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.

本题的巧妙就在于作为一个BST，如果要找两个节点的公共祖先节点，那么这个祖先节点的key值一定介于两个节点的key值之间。

#include <iostream>
#include <vector>
#include <set>
using namespace std;
int main(){
	int n,m;
	scanf("%d %d",&m,&n);
	vector<int>pre(n);
	set<int>s;
	for(int i=0;i<n;i++){
		scanf("%d",&pre[i]);
		s.insert(pre[i]);
	}
	for(int i=0;i<m;i++){
		int u,v,temp;
		scanf("%d %d",&u,&v);
		if(s.find(u)==s.end()&&s.find(v)==s.end())
			printf("ERROR: %d and %d are not found.\n",u,v);
		else if(s.find(u)==s.end()||s.find(v)==s.end())
			printf("ERROR: %d is not found.\n", s.find(u)==s.end()?u:v);
		else{
			for(int j=0;j<n;j++){
				temp=pre[j];
				if((temp<=u&&temp>=v)||(temp>=u&&temp<=v))
                	break;
			}
			if(temp==u||temp==v)
				printf("%d is an ancestor of %d.\n",temp,temp==u?v:u);
			else
				printf("LCA of %d and %d is %d.\n", u, v, temp);
		}
	}
}