A1143 Lowest Common Ancestor (30 point(s))

二叉查找树的共同祖先

1. 原文

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.

2. 解析

不同于A1155 该树是二叉查找树 左< 根< 右 先序先输出的必是根

直接比较根的值，当u，v 值在根两侧时即为答案

for (int i = 0; i < n; ++i){
    a=pre[i];
    if ((u<=a&&v>=a)||(u>=a&&v<=a)){
        break;
     }	
}

3. AC代码

#include<cstdio>
#include<map>
using namespace std;
map<int,int> exist;
int pre[10010]={};
int main(){
	int m,n;
	scanf("%d%d",&m,&n);
	for (int i = 0; i < n; ++i)
	{
		scanf("%d",&pre[i]);
		exist[pre[i]]++;
	}
	int u,v;
	while(m--)
	{
		scanf("%d%d",&u,&v);
		int a=0;
        for (int i = 0; i < n; ++i)
        {
        	a=pre[i];
        	if ((u<=a&&v>=a)||(u>=a&&v<=a))
        	{
        		break;
        	}	
        }
		if (exist[u]==0&&exist[v]==0)
		{
			printf("ERROR: %d and %d are not found.\n",u,v);
		}else if(exist[u]==0){
			printf("ERROR: %d is not found.\n",u);
		}else if(exist[v]==0){
			printf("ERROR: %d is not found.\n",v);
		}else if(u==a){
			printf("%d is an ancestor of %d.\n",u,v);
		}else if(v==a){
			printf("%d is an ancestor of %d.\n",v,u);
		}else{
			printf("LCA of %d and %d is %d.\n",u,v,a);
		}
	}
	return 0;
}