1143. Lowest Common Ancestor (30)

1143. Lowest Common Ancestor (30)
时间限制 200 ms 内存限制 65536 kB 代码长度限制 16000 B
判题程序 Standard 作者 CHEN, Yue

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 (<= 1000), the number of pairs of nodes to be tested; and N (<= 10000), 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.

注意点：不能直接进行插入建树，会超时…

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <algorithm>
#include <stack>
#include <set>
using namespace std;
const int Maxn = 10000;

typedef struct node {
    int data;
    struct node* lchild;
    struct node* rchild;
    node() { lchild = rchild = nullptr; }
}TNode, *PTNode;


int M, N;
set<int> keys;
int pre[Maxn], in[Maxn];

//void Insert(PTNode  & root, int data) {
//  if (!root) {
//      root = new TNode;
//      root->data = data;
//      return;
//  }
//  else if (root->data < data) Insert(root->rchild, data);
//  else if (root->data > data) Insert(root->lchild, data);
//}

PTNode CreateBST(int pre[], int in[], int p_l, int p_r, int i_l, int i_r) {
    if (p_l > p_r) return nullptr;
    PTNode root = new TNode;
    root->data = pre[p_l];

    int idx = i_l;
    while (idx < i_r && in[idx] != pre[p_l])++idx;
    root->lchild = CreateBST(pre, in, p_l + 1, p_l + idx - i_l - 1, i_l, idx - 1);
    root->rchild = CreateBST(pre, in, p_l + idx - i_l + 1, p_r, idx + 1, i_r);
    return root;
}


int findLCA(PTNode root, int e1, int e2, bool &ans1, bool &ans2) {
    if (e1 < root->data && e2 < root->data)  return findLCA(root->lchild, e1, e2, ans1, ans2);//如果当前根节点均小于查询的两个节点，则最小近公共祖先在左子树
    else if (e1 > root->data && e2 > root->data)  return findLCA(root->rchild, e1, e2, ans1, ans2);//如果当前根节点均大于查询的两个节点，则最小近公共祖先在右子树
    else if (e1 > root->data && e2 < root->data || e1 < root->data && e2 > root->data) return root->data;//如果一大一小，那么当前节点肯定是其最近公共祖先
    else { //否则其中一个值一定是另一个值的祖先
        if (e1 != root->data)ans2 = true;
        else ans1 = true;
        return root->data;
    }
}


int main()
{
#ifdef _DEBUG
    freopen("data.txt", "r+", stdin);
#endif

    std::ios::sync_with_stdio(false);
    int data;
    cin >> M >> N;
    for (int i = 0; i<N; ++i) {
        cin >> data;
        in[i] = pre[i] = data;
        keys.insert(data);
    }
    sort(in, in + N);
    PTNode root = CreateBST(pre, in, 0, N - 1, 0, N - 1);

    while (M--) {
        bool f1flag = true, f2flag = true;
        int e1, e2;
        cin >> e1 >> e2;
        if (keys.find(e1) == keys.end()) f1flag = false;
        if (keys.find(e2) == keys.end()) f2flag = false;

        if (f1flag == false || f2flag == false) {
            if (f1flag && !f2flag) cout << "ERROR: " << e2 << " is not found.\n";
            else if (!f1flag && f2flag) cout << "ERROR: " << e1 << " is not found.\n";
            else cout << "ERROR: " << e1 << " and " << e2 << " are not found.\n";
            continue;
        }
        bool ans1 = false, ans2 = false;
        int LCAD = findLCA(root, e1, e2, ans1, ans2);
        if (ans1)cout << e1 << " is an ancestor of " << e2 << ".\n";
        else if (ans2) cout << e2 << " is an ancestor of " << e1 << ".\n";
        else cout << "LCA of " << e1 << " and " << e2 << " is " << LCAD << ".\n";
    }

    return 0;
}