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  • 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

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  • 原文地址:https://www.cnblogs.com/zlrrrr/p/10336975.html
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