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 componentswhere 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
1 #include <iostream> 2 #include <vector> 3 #include<set> 4 using namespace std; 5 vector<vector<int>>G; 6 int N, maxH = 0; 7 bool visit[10010]; 8 set<int>res; 9 vector<int>temp; 10 11 void DFS(int node, int H) 12 { 13 if (H > maxH) 14 { 15 temp.clear(); 16 temp.push_back(node);//更新新的根节点 17 maxH = H; 18 } 19 else if (H == maxH) 20 temp.push_back(node);//相同的最优解 21 visit[node] = true; 22 for (int i = 0; i < G[node].size(); ++i) 23 if (visit[G[node][i]] == false) 24 DFS(G[node][i], H + 1); 25 } 26 27 int main() 28 { 29 int a, b, s1 = 0, cnt = 0; 30 cin >> N; 31 G.resize(N+1); 32 for (int i = 1; i < N; ++i) 33 { 34 cin >> a >> b; 35 G[a].push_back(b); 36 G[b].push_back(a); 37 } 38 for (int i = 1; i <= N; ++i) 39 { 40 if (visit[i] == false)//开始深度搜索遍历，如果是一个联通区域，则只会执行一次 41 { 42 DFS(i, 1); 43 if (i == 1) 44 { 45 if (temp.size() != 0) 46 s1 = temp[0]; 47 for (int j = 0; j < temp.size(); ++j) 48 res.insert(temp[j]); 49 } 50 cnt++;//计算集合数 51 } 52 } 53 if (cnt != 1) 54 printf("Error: %d components ", cnt); 55 else 56 { 57 temp.clear(); 58 maxH = 0; 59 fill(visit, visit + N + 1, false); 60 DFS(s1, 1); 61 for (int j = 0; j < temp.size(); ++j) 62 res.insert(temp[j]); 63 for (auto r : res) 64 cout << r << endl; 65 } 66 return 0; 67 }