A graph which is connected and acyclic can be considered a tree. The hight 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

Sample Output 2:

Error: 2 components

 先利用並查集求有幾個集合。。

然後用兩遍dfs。。

第一遍dfs任意一個點，求出最大深度點的集合x。。。

第二遍dfs最大深度的任意一個點，求出另一個最大深度點的集合y。。

求兩個集合的並集就是題目答案。。。

程式碼如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;
const int maxn=10005;
int n;
int flag=0;
int vis[maxn];
int len[maxn];
int is[maxn];
int a[maxn];
int Lenn=0;
vector<int>ve[maxn];
int Find(int x)
{
    if(x==a[x])
        return x;
    return a[x]=Find(a[x]);
}
void unit(int x,int y)
{
    int fx=Find(x);
    int fy=Find(y);
    if(fx!=fy)
        a[fx]=fy;
}
void init()
{
    memset(vis,0,sizeof(vis));
    memset(len,0,sizeof(len));
    memset(is,0,sizeof(is));
}
void dfs (int x,int height)
{
    Lenn=max(Lenn,height);
    vis[x]=1;
    len[x]=height;
    for (int i=0;i<ve[x].size();i++)
    {
        int v=ve[x][i];
        if(!vis[v])
        {
            vis[v]=1;
            dfs(v,height+1);
        }
    }
}
int main()
{
    scanf("%d",&n);
    for (int i=1;i<=n;i++)
        a[i]=i;
    for (int i=1;i<n;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        unit(x,y);
        ve[x].push_back(y);
        ve[y].push_back(x);
    }
    for (int i=1;i<=n;i++)
         if(a[i]==i)
            flag++;
    if(flag>1)
        printf("Error: %d components\n",flag);
    else
    {
        init();
        dfs(1,0);
        int x;
        for (int i=1;i<=n;i++)
            if(len[i]==Lenn)
            {
                is[i]=1;
                x=i;
            }
        memset (vis,0,sizeof(vis));
        memset (len,0,sizeof(len));
        Lenn=0;
        dfs(x,0);
        for (int i=1;i<=n;i++)
            if(len[i]==Lenn)
                is[i]=1;
        for (int i=1;i<=n;i++)
            if(is[i])
               printf("%d\n",i);
    }
    return 0;
}