题干：

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

Sample Output 2:

Error: 2 components

题目大意：

一个连通的非循环图可以看作是一个树。树的高度取决于所选的根。现在你应该找到根，结果是最高的树。这样的根叫做最深的根。对于每个测试用例，在一行中打印每个最深的根。如果这样的根不是惟一的，则按其数字的递增顺序打印它们。如果给定的图形不是树，则打印Error: K components，其中K是图中连图分量的数量。

解题报告：

  因为时限是两秒，直接枚举每一个点当根节点就可以。但是这题其实可以优化，只枚举叶子节点就可以。甚至可以用树的直径去乱搞一波，时间复杂度会更优。但这题时限很宽，所以没必要优化。

AC代码： 

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
int n,m,f[MAX],rt[MAX],dep[MAX];
vector<int> vv[MAX],ans;
int getf(int v) {return f[v] == v ? v : f[v] = getf(f[v]);
}
void merge(int u,int v) {int t1 = getf(u),t2 = getf(v);f[t2] = t1;
}
int mx;
void dfs(int u,int fa) {dep[u] = dep[fa] + 1;mx = max(mx,dep[u]);int up = vv[u].size();for(int i = 0; i<up; i++) {int v = vv[u][i];if(v == fa) continue;dfs(v,u);}
}
int main()
{cin>>n;for(int i = 1; i<=n; i++) f[i] = i;for(int u,v,i = 1; i<=n-1; i++) cin>>u>>v,vv[u].pb(v),vv[v].pb(u),merge(u,v);int cnt = 0;for(int i = 1; i<=n; i++) {if(f[i] == i) cnt++;}if(cnt != 1) {printf("Error: %d components\n",cnt);return 0 ;}for(int root = 1; root<=n; root++) {mx = 0;dfs(root,0);rt[root] = mx;}mx = 0;for(int i = 1; i<=n; i++) {if(rt[i] > mx) {mx = rt[i];ans.clear();ans.pb(i);}else if(rt[i] == mx) ans.pb(i);} sort(ans.begin(),ans.end());for(auto x : ans) {printf("%d\n",x);}return 0 ;
}