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 (≤) 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题意：
一个图可以看成是一棵树，这颗树的高度取决于所选的根节点。输出使树的高度最高的根节点。如果所给出的图不能够用树来表示，则输出图中连通分量的个数。
思路：
用DFS深搜找出连通分量的个数，第一次深搜找出以一号结点作为根节点的深度最大的那个结点，然后再一次结点进行第二次深搜，找出深度最深的结点。
Code：
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 int n; 6 vector<int> ans; 7 vector<bool> visited; 8 vector<vector<int> > grap; 9 int max_deep = 0; 10 11 void DFS(int node, int level) { 12 if (max_deep < level) { 13 ans.clear(); 14 ans.push_back(node); 15 max_deep = level; 16 } else if (max_deep == level) { 17 ans.push_back(node); 18 } 19 visited[node] = true; 20 for (int it : grap[node]) { 21 if (visited[it] == false) { 22 DFS(it, level + 1); 23 } 24 } 25 } 26 27 int main() { 28 int v1, v2; 29 cin >> n; 30 grap.resize(n + 1); 31 visited.resize(n + 1, false); 32 for (int i = 1; i < n; ++i) { 33 cin >> v1 >> v2; 34 grap[v1].push_back(v2); 35 grap[v2].push_back(v1); 36 } 37 int count = 0, s1 = 0; 38 set<int> s; 39 for (int i = 1; i <= n; ++i) { 40 if (visited[i] == false) { 41 DFS(i, 1); 42 if (i == 1) { 43 if (ans.size() != 0) s1 = ans[0]; 44 for (int it : ans) s.insert(it); 45 } 46 count++; 47 } 48 } 49 50 if (count > 1) { 51 cout << "Error: " << count << " components" << endl; 52 } else { 53 ans.clear(); 54 max_deep = 0; 55 visited.clear(); 56 visited.resize(n + 1, false); 57 DFS(s1, 1); 58 for (int it : ans) s.insert(it); 59 for (int it : s) cout << it << endl; 60 } 61 return 0; 62 }
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