PAT 1151 LCA in a Binary Tree (30)

1151 LCA in a Binary Tree (30) （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.

Given any two nodes in a binary tree, 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 binary tree, respectively. In each of the following two lines, N distinct integers are given as the inorder and preorder traversal sequences of the binary tree, respectively. It is guaranteed that the binary tree can be uniquely determined by the input sequences. 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 binary tree, 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
7 2 3 4 6 5 1 8
5 3 7 2 6 4 8 1
2 6
8 1
7 9
12 -3
0 8
99 99

Sample Output:

LCA of 2 and 6 is 3.
8 is an ancestor of 1.
ERROR: 9 is not found.
ERROR: 12 and -3 are not found.
ERROR: 0 is not found.
ERROR: 99 and 99 are not found.

思路：

不需要特地弄个树形结构然后建树，可以直接根据先序遍历和中序遍历得出每个节点的父节点和深度。需要注意的是测试的数据不是按照样例这样说8个节点然后编号就是1-8，很有可能有奇怪的数字，我一开始只用数组就出现了一个段错误，后来全改成map又发现超时，所以需要自己手动编个号。

代码：

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cctype>
#include <climits>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>  
#include <set>
using namespace std;

map<int, int> mp1, mp2;     //mp1-节点权值转手动编号 mp2-手动编号转权值
int pre[10005], in[10005];  //pre-先序遍历 in-中序遍历
int f[10005], d[10005];     //f-父节点 d-深度

void build(int* pre, int* in, int len, int father, int depth)
{
	if (len == 0)
		return;
	int i = 0;
	while (pre[0] != in[i])
		i++;
	build(pre + 1, in, i, pre[0], depth + 1);
	build(pre + i + 1, in + i + 1, len - i - 1, pre[0], depth + 1);
	f[mp1[pre[0]]] = mp1[father];
	d[mp1[pre[0]]] = depth;
}

int LCA(int a, int b)
{
	if (d[a] > d[b])
		swap(a, b);
	while (d[b] > d[a])
		b = f[b];
	while (a != b)
	{
		a = f[a];
		b = f[b];
	}
	return a;
}

int main()
{
	int m, n;
	scanf("%d%d", &m, &n);
	for (int i = 0; i < n; i++)
	{
		scanf("%d", &in[i]);
		mp1[in[i]] = i + 1;
		mp2[i + 1] = in[i];
	}
	for (int i = 0; i < n; i++)
		scanf("%d", &pre[i]);
	build(pre, in, n, pre[0], 1);
	for (int i = 0; i < m; i++)
	{
		int a, b, cnt = 0;
		scanf("%d%d", &a, &b);
		if (mp1[a] == 0 && mp1[b] == 0)
			printf("ERROR: %d and %d are not found.\n", a, b);
		else if (mp1[a] == 0)
			printf("ERROR: %d is not found.\n", a);
		else if (mp1[b] == 0)
			printf("ERROR: %d is not found.\n", b);
		else
		{
			int ans = LCA(mp1[a], mp1[b]);
			if (mp2[ans] == a)
				printf("%d is an ancestor of %d.\n", a, b);
			else if (mp2[ans] == b)
				printf("%d is an ancestor of %d.\n", b, a);
			else
				printf("LCA of %d and %d is %d.\n", a, b, mp2[ans]);
		}
	}
	return 0;
}