zoukankan      html  css  js  c++  java
  • PAT 1143 Lowest Common Ancestor

    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 <iostream>
    #include <vector>
    #include <map>
    using namespace std;
    map<int, bool> mp;
    int main() {
        int m, n, u, v, a;
        scanf("%d %d", &m, &n);
        vector<int> pre(n);
        for (int i = 0; i < n; i++) {
            scanf("%d", &pre[i]);
            mp[pre[i]] = true;
        }
        for (int i = 0; i < m; i++) {
            scanf("%d %d", &u, &v);
            for(int j = 0; j < n; j++) {
                a = pre[j];
                if ((a > u && a < v)|| (a > v && a < u) || (a == u) || (a == v)) break;
            } 
            if (mp[u] == false && mp[v] == false)
                printf("ERROR: %d and %d are not found.
    ", u, v);
            else if (mp[u] == false || mp[v] == false)
                printf("ERROR: %d is not found.
    ", mp[u] == false ? u : v);
            else if (a == u || a == v)
                printf("%d is an ancestor of %d.
    ", a, a == u ? v : u);
            else
                printf("LCA of %d and %d is %d.
    ", u, v, a);
        }
        return 0;
    }
    
  • 相关阅读:
    CFileDialog打开多个文件失败 返回错误 FNERR_BUFFERTOOSMALL
    VC 控件集合
    [摘]思科认证三步走及找工作的七大职业走向
    windows无法配置此无线连接解决办法
    VS 2019 项目添加引用,提示:对COM组件的调用返回了错误HRESULT E_FAIL
    c# winform 获取当前程序运行根目录
    DataGridView 转换成 DataTable
    SQL Server日志文件过大 大日志文件清理方法 不分离数据库
    行动吧!让自己骚起来
    抖音很火的3D旋转相册 (源代码分享)
  • 原文地址:https://www.cnblogs.com/A-Little-Nut/p/9652283.html
Copyright © 2011-2022 走看看