1143 Lowest Common Ancestor (30 分)

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.

这题可以建树做，因为搜索树的话，它的中序遍历就是排好序的。
但是这题可以用更巧妙的方法做，将图画出来，可以看出来，如果ans是x和y的最近公共祖先
那么ans是介于x和y之间的【min(x,y), max(x,y)】

#include <bits/stdc++.h>
using namespace std;
int n,m,x,y,ans,an[100005];
map<int,int> mp;
int main(){
    cin >> n >> m;
    for(int i = 1; i <= m; i++){
        cin >> an[i];
        mp[an[i]] = 1;
    }
    for(int i = 0 ; i < n; i++){
        cin >> x >> y;
        for(int j = 1; j<= m; j++){
            ans = an[j];
            if((ans >= x&&ans <= y)||(ans>=y && ans <= x))
                break;
        }
        if(mp[x] == 0 && mp[y] == 0)
            printf("ERROR: %d and %d are not found.\n", x,y);
        else if(mp[x] == 0)
            printf("ERROR: %d is not found.\n", x);
        else if(mp[y] == 0)
            printf("ERROR: %d is not found.\n", y);
        else if(ans==x || ans==y)
            printf("%d is an ancestor of %d.\n", ans==x?x:y,ans==x?y:x);
        else
            printf("LCA of %d and %d is %d.\n", x,y,ans);
    }
    return 0;
}