1021 Deepest Root (25)（25 分）

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

 思路：

輸入一個無向圖，首先判斷是不是樹。n個點n-1條邊的無向圖如果是連通圖就一定是樹，如果不連通就勢必有環，但凡有一個環就會多一個連通分支，所以只要求出環的個數，連通分支數就是環的個數+1。然後如果是樹，這個問題就成了求樹的直徑（最長的簡單路），選取任意一點a先做一遍dfs，求出距離該點最遠的點集合A，再從該集合中任選一點b再做一遍dfs，求出距離最遠的點集合B，答案就是集合A、B的並集。

程式碼：

#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;

vector<vector<int>> g;
vector<int> v;
set<int> s;
int vis[10005], maxd = 0;

void dfs(int x, int depth)
{
	vis[x] = 1;
	if (depth > maxd)
	{
		maxd = depth;
		v.clear();
		v.push_back(x);
	}
	else if (depth == maxd)
		v.push_back(x);
	for (int i = 0; i < g[x].size(); i++)
	{
		if (!vis[g[x][i]])
			dfs(g[x][i], depth + 1);
	}
}

int main()
{
	int n, cnt = 1;
	cin >> n;
	g.resize(n + 1);
	memset(vis, 0, sizeof(vis));
	for (int i = 1; i < n; i++)
	{
		int from, to;
		cin >> from >> to;
		g[from].push_back(to);
		g[to].push_back(from);
		if (vis[from] && vis[to])
			cnt++;
		else
			vis[from] = vis[to] = 1;
	}
	if (cnt > 1)
		cout << "Error: " << cnt << " components" << endl;
	else
	{
		memset(vis, 0, sizeof(vis));
		dfs(1, 1);
		for (int i = 0; i < v.size(); i++)
			s.insert(v[i]);
		v.clear();
		maxd = 0;
		memset(vis, 0, sizeof(vis));
		dfs(*s.begin(), 1);
		for (int i = 0; i < v.size(); i++)
			s.insert(v[i]);
		for (set<int>::iterator iter = s.begin(); iter != s.end(); iter++)
			cout << *iter << endl;
	}
	return 0;
}