PAT甲级A1143 Lowest Common Ancestor （30 分)

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.

题意：给出一棵BST树的先序序列，再给出m组顶点u，v，让你找出u和v的最近公共祖先，并按要求输出。

思路：如果先建立BST树在通过DFS求最近公共祖先，则会有几组case超时，考虑到这是一棵BST树，且给定的是先序序列，有如下特点：对于任意给定的顶点u和v，从头开始遍历该先序序列，如果当前顶点a>=u并且a<=v或者a>=v且a<=u，则可以断定a是u和v的最近公共祖先，为了防止u或v不在序列中的情况，可以使用map在输入时，做一个映射，如果u或v不在map中则不是树中的顶点。

参考代码：

#include<cstdio>
#include<unordered_map>
using namespace std;
unordered_map<int,int> mp;
int n,m,u,v,a[10010];
int main()
{
	scanf("%d%d",&m,&n);
	for(int i=0;i<n;i++){
		scanf("%d",&a[i]);
		mp[a[i]]=1;
	}
	for(int k=0;k<m;k++){
		scanf("%d%d",&u,&v);
		if(mp.find(u)==mp.end()&&mp.find(v)==mp.end())
			printf("ERROR: %d and %d are not found.\n",u,v);
		else if(mp.find(u)==mp.end())
			printf("ERROR: %d is not found.\n",u);
		else if(mp.find(v)==mp.end())
			printf("ERROR: %d is not found.\n",v);
		else{
			bool flag=false;
			int LCA=0;
			for(int i=0;i<n&&!flag;i++){
				if((a[i]>=u&&a[i]<=v)||(a[i]>=v&&a[i]<=u)){
					LCA=a[i];
					flag=true;
				}
			}
			if(LCA==u) printf("%d is an ancestor of %d.\n",u,v);
			else if(LCA==v) printf("%d is an ancestor of %d.\n",v,u);
			else printf("LCA of %d and %d is %d.\n",u,v,LCA);
		}
	}
	return 0;
}