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



 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\n", 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 }

 

posted @ 2019-07-23 10:51  自由之翼Az  阅读(184)  评论(0)  编辑  收藏  举报