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 1:5 1 2 1 3 1 4 2 5Sample Output 1:
3 4 5Sample Input 2:
5 1 3 1 4 2 5 3 4Sample 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 <iostream>
#include <stdio.h>
#include <queue>
#include <set>
using namespace std;
#define MAX 10002
bool visited[MAX];
vector<vector<int> > 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<int> 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<int> root;
for(int i = 1; i<= N; i++){
root.insert(FindSet(i));
}
if(root.size() != 1){
printf("Error: %d components
",root.size());
return 0;
}
vector<int> 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
",maxDeepNodes[i]);
}
return 0;
}