题目如下：

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 (<=10000) 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:

51 21 31 42 5

Sample Output 1:

345

Sample Input 2:

51 31 42 53 4

Sample Output 2:

Error: 2 components

这个题有N个顶点，N-1条边，属于稀疏图，应该用邻接矩阵而不是邻接表，否则会内存超限。

刚开始拿到这道题目的时候我想到的方法是用带颜色标记的BFS判断环路，并且排除掉，但是这样会超时，后来在网上查阅，看到大家都是使用并查集来判断是否是树，通过把所有元素归入并查集，并且在查询根的时候压缩路径，如果所有元素都是同根的，那么把他们加入Set后集合的规模为1，说明是一棵完整的树，如果不是如此，则说明图由多个连通的部分组成，连通部分的个数等于根的个数，因此只需要输出Set的规模即可。

如果是一棵树，只需要使用BFS来遍历树中的每一个顶点，来计算深度。

为了计算深度，改进BFS，加入三个变量：head、last、tail，last表示这一层的最后一个元素，tail表示下一层的某个元素，初始化时，让last=s，代表第一层只有源点，接下来在出队一个结点，并让head等于这个结点，然后在遍历结点的邻接点时不断的更新tail，不难看出tail在最后一次更新后指向的是下一层的最后一个元素，然后判断head是否等于last，如果是代表本层结束，把层计数变量+1，更新last为tail，以此类推即可。

#include 
    
     #include 
     
      #include 
      
       #include 
       
        using namespace std;#define MAX 10002bool visited[MAX];vector
        
         
           > Graph;int N;int P[MAX];void initSet(int N){ for(int i = 1; i <= N; i++) P[i] = i;}void CompressSet(int x, int top){ if(P[x] != top){ CompressSet(P[x],top); P[x] = top; }}int FindSet(int x){ if(x != P[x]){ int t = FindSet(P[x]); CompressSet(x,t); } return P[x];}void UnionSet(int x, int y){ int p1 = FindSet(x); int p2 = FindSet(y); P[p1] = p2;}int BFS(int s){ for(int i = 1; i <= N; i++) visited[i] = false; int head,tail,last; int stage = 0; queue
          
            Q; Q.push(s); visited[s] = 1; last = s; while(!Q.empty()){ int v = Q.front(); visited[v] = 1; head = v; Q.pop(); for(int index = 0; index < Graph[v].size(); index++){ int w = Graph[v][index]; if(!visited[w]){ Q.push(w); tail = w; } } if(head == last){ last = tail; stage++; } } return stage;}int main(){ cin >> N; int v1,v2; Graph.resize(N + 1); initSet(N); for(int i = 1; i <= N; i++){ visited[i] = false; } for(int i = 1; i < N; i++){ scanf("%d%d",&v1,&v2); Graph[v1].push_back(v2); Graph[v2].push_back(v1); } for(int i = 1; i <= N; i++){ for(int j = 0; j < Graph[i].size(); j++){ UnionSet(i,Graph[i][j]); } } set
           
             root; for(int i = 1; i<= N; i++){ root.insert(FindSet(i)); } if(root.size() != 1){ printf("Error: %d components\n",root.size()); return 0; } vector
            
              maxDeepNodes; maxDeepNodes.clear(); int maxDepth = 0; int depth = 0; for(int i = 1; i <= N; i++){ depth = BFS(i); if(depth > maxDepth){ maxDepth = depth; maxDeepNodes.clear(); maxDeepNodes.push_back(i); }else if(depth == maxDepth){ maxDeepNodes.push_back(i); } } for(int i = 0; i < maxDeepNodes.size(); i++){ printf("%d\n",maxDeepNodes[i]); } return 0;}