PAT甲级——A1021 Deepest Root

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

 #include <iostream>
 #include <vector>
 #include<set>
 using namespace std;
 vector<vector<int>>G;
 int N, maxH = ;
 bool visit[];
 set<int>res;
 vector<int>temp;

 void DFS(int node, int H)
 {
     if (H > maxH)
     {
         temp.clear();
         temp.push_back(node);//更新新的根节点
         maxH = H;
     }
     else if (H == maxH)
         temp.push_back(node);//相同的最优解
     visit[node] = true;
     for (int i = ; i < G[node].size(); ++i)
         if (visit[G[node][i]] == false)
             DFS(G[node][i], H + );
 }

 int main()
 {
     int a, b, s1 = , cnt = ;
     cin >> N;
     G.resize(N+);
     for (int i = ; i < N; ++i)
     {
         cin >> a >> b;
         G[a].push_back(b);
         G[b].push_back(a);
     }
     for (int i = ; i <= N; ++i)
     {
         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
         {
             DFS(i, );
             if (i == )
             {
                 if (temp.size() != )
                     s1 = temp[];
                 for (int j = ; j < temp.size(); ++j)
                     res.insert(temp[j]);
             }
             cnt++;//计算集合数
         }
     }
     if (cnt != )
         printf("Error: %d components\n", cnt);
     else
     {
         temp.clear();
         maxH = ;
         fill(visit, visit + N + , false);
         DFS(s1, );
         for (int j = ; j < temp.size(); ++j)
             res.insert(temp[j]);
         for (auto r : res)
             cout << r << endl;
     }
     return ;
 }