PAT 甲级 1143 Lowest Common Ancestor

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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.\n", a, b);
        else if(!mp[a] || !mp[b])
            printf("ERROR: %d is not found.\n", 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.\n", a, b);
            else if(root == b)
                printf("%d is an ancestor of %d.\n", b, a);
            else printf("LCA of %d and %d is %d.\n", 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

PAT 甲级 1143 Lowest Common Ancestor

标签：sed   c++   rop   erro   name   can   other   namespace   set   

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