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

    https://pintia.cn/problem-sets/994805342720868352/problems/994805343727501312

    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.

    代码:

    #include <bits/stdc++.h>
    using namespace std;
    
    int N, M;
    vector<int> pre;
    map<int, int> mp;
    
    int main() {
        scanf("%d%d", &M, &N);
        pre.resize(N);
        for(int i = 0; i < N; i ++) {
            scanf("%d", &pre[i]);
            mp[pre[i]] = 1;
        }
        while(M --) {
            int a, b;
            scanf("%d%d", &a, &b);
            if(!mp[a] && !mp[b])
                printf("ERROR: %d and %d are not found.
    ", a, b);
            else if(!mp[a] || !mp[b])
                printf("ERROR: %d is not found.
    ", mp[a] ? b : a);
            else {
                int root;
                for(int i = 0; i < N; i ++) {
                    root = pre[i];
                    if((root >= a && root <= b) || (root <= a && root >= b)) break;
                }
    
                if(root == a)
                    printf("%d is an ancestor of %d.
    ", a, b);
                else if(root == b)
                    printf("%d is an ancestor of %d.
    ", b, a);
                else printf("LCA of %d and %d is %d.
    ", a, b, root);
            }
        }
        return 0;
    }
    

      用 mp 记下是否出现过 然后只要从头找满足值在 a b 之间的就是根了 

    FHFHFH 过年之前最后一个工作日

  • 相关阅读:
    保持URL不变和数字验证
    centOS ftp key?
    本地环境测试二级域名
    linux 解决You don't have permission to access 问题
    php smarty section loop
    php header Cannot modify header information headers already sent by ... 解决办法
    linux部分命令
    Linux 里面的文件操作权限说明
    用IT网络和安全专业人士视角来裁剪云的定义
    SQL Server 2008 R2炫酷报表"智"作有方
  • 原文地址:https://www.cnblogs.com/zlrrrr/p/10336975.html
Copyright © 2011-2022 走看看