PAT甲级 1143

1143 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 :给出二叉排序树的结点数目和前序遍历序列,再给出数对结点,找出他们的最近公共祖先,根据实际情况按照题目要求的格式进行输出。

#include <bits/stdc++.h>
using namespace std;
const int maxn = 10001;
int a[maxn];
int n, m;
unordered_set<int> S;
unordered_set<int>::iterator it1, it2;
int work(int now) {
	int mid, l = now+1, r = n;
	while(l <= r) {
		mid = (l+r) / 2;
		if(a[mid] < a[now]) {
			l = mid+1;
		}
		else {
			r = mid-1;
		}
	}
	return l;
}
int main() {
	scanf("%d %d", &m, &n); S.clear();
	for(int i = 1; i <= n; ++i) {
		scanf("%d", &a[i]);
		S.insert(a[i]);
	}
	while(m--) {
		int u, v;
		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()) {
			printf("ERROR: %d is not found.\n", u);
		}
		else if(S.find(v) == S.end()) {
			printf("ERROR: %d is not found.\n", v);
		}
		else {
			int now = 1, mins = min(u, v), maxx = max(u, v);
			while(true) {
				if(a[now] < mins) {
					now = work(now);
				}
				else if(a[now] > maxx) {
					++now;
				}
				else {
					break;
				}
			}
			if(a[now] == u) {
				printf("%d is an ancestor of %d.\n", u, v);
			}
			else if(a[now] == v) {
				printf("%d is an ancestor of %d.\n", v, u);
			}
			else {
				printf("LCA of %d and %d is %d.\n", u, v, a[now]);
			}
		}
	}
	return 0;
}

 

### 关于 PAT 甲级 1024 题目 PAT (Programming Ability Test) 是一项编程能力测试,其中甲级考试面向有一定编程基础的学生。对于 PAT 甲级 1024 题目,虽然具体题目描述未直接给出,但从相似类型的题目分析来看,这类题目通常涉及较为复杂的算法设计。 #### 数据结构的选择与实现 针对此类问题,常用的数据结构包括但不限于二叉树节点定义: ```cpp struct Node { int val; Node* lchild, *rchild; }; ``` 此数据结构用于表示二叉树中的节点[^1]。通过这种方式构建的二叉树能够支持多种遍历操作,如前序、中序和后序遍历等。 #### 算法思路 当处理涉及到图论的问题时,深度优先搜索(DFS)是一种常见的解题策略。特别是当需要寻找最优路径或访问尽可能多的节点时,结合贪心算法可以在某些情况下提供有效的解决方案[^2]。 #### 输入输出格式说明 根据以往的经验,在解决 PAT 类型的问题时,输入部分往往遵循特定模式。例如,给定 N 行输入来描述每个节点的信息,每行按照如下格式:“Address Data Next”,这有助于理解如何解析输入并建立相应的数据模型[^4]。 #### 数学运算示例 有时也会遇到基本算术表达式的求值问题,比如分数之间的加减乘除运算。下面是一些简单的例子展示不同情况下的计算结果: - \( \frac{2}{3} + (-2) = -\frac{7}{3}\) -2) = -\frac{4}{3}\) - \( \frac{2}{3} ÷ (-2) = -\frac{1}{3}\) 这些运算是基于样例提供的信息得出的结果[^3]。
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