1021 Deepest Root(甲級)

1021 Deepest Root (25分)
A graph which is connected and acyclic can be considered a tree. The height of the tree depends on the selected root. Now you are supposed to find the root that results in a highest tree. Such a root is called the deepest root.
Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (≤10
​4
​​ ) which is the number of nodes, and hence the nodes are numbered from 1 to N. Then N−1 lines follow, each describes an edge by given the two adjacent nodes’ numbers.
Output Specification:

For each test case, print each of the deepest roots in a line. If such a root is not unique, print them in increasing order of their numbers. In case that the given graph is not a tree, print Error: K components where K is the number of connected components in the graph.
Sample Input 1:

5
1 2
1 3
1 4
2 5
Sample Output 1:

3
4
5
Sample Input 2:

5
1 3
1 4
2 5
3 4

#include<iostream>
#include<vector>
#include<memory.h>
using namespace std;
int n;
const int maxn = 10001;
vector<int>v[maxn];
bool vist[maxn]{ false };
int max_depth = -1;//統計以任意節點作為根的最大深度，也就是該顆樹的最大深度
int max_tmp = -1;//記錄所有節點的最大深度，所有樹中的最大深度
void dfs(int root, int depth)
{
	if (max_depth < depth)//求最大深度
	{
		max_depth = depth;
	}
	vist[root] = true;
	for (int i = 0; i < v[root].size(); i++)
	{
		if (!vist[v[root][i]])
		{
			dfs(v[root][i], depth + 1);
		}
	}
}
int main()
{
	cin >> n;
	vector<int>count;
	for (int i = 0; i < n - 1; i++)
	{
		int value1, value2;
		cin >> value1 >> value2;
		v[value1].push_back(value2);
		v[value2].push_back(value1);
	}
	for (int i = 1; i <= n; i++)
	{
		int cnt = 0;
		max_depth = 0;
		memset(vist, false, maxn);
		if (i == 1)//先判斷是否是聯通的
		{
			for (int j = 1; j <= n; j++)
			{
				if (!vist[j])
				{
					cnt++;
					dfs(j, 1);
				}
			}
			if (cnt >= 2)
			{
				cout << "Error: " << cnt << " components";
				return 0;
			}
		}
		else
		{
			dfs(i, 1);
		}
		if (max_tmp < max_depth)//更新最大深度，並將根節點入vector
		{
			count.clear();
			max_tmp = max_depth;
			count.push_back(i);
		}
		else if (max_tmp == max_depth)//相等，根節點入vector
		{
			count.push_back(i);
		}
	}
	for (int i = 0; i < count.size(); i++)
	{
		cout << count[i] << endl;
	}
}