题目:

A graph which is connected and acyclic can be considered a tree. The hight 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

分析:

DFS。参考博客：点击前往。但依然有两组数据出错。

代码:

 1 #include<iostream>
 2 #include<cstring>
 3 #include<vector>
 4 #include<algorithm>
 5 using namespace std;
 6 struct Node{
 7     vector<int> points;
 8 }nodes[10001];
 9 
10 bool visited[10001];
11 int n;
12 int height=0,components=0;
13 vector<int> farthestNode;
14 
15 void dfs(int root,int level){
16     visited[root]=true;
17     if(level>height){        //若高度更高，则原本的集合清空，将最高的更新 
18         height=level;
19         farthestNode.clear();
20         farthestNode.push_back(root);
21     }else if(level==height){    //同等高度，将该结点添加进去 
22         farthestNode.push_back(root);
23     }
24     for(int i=0;i<nodes[root].points.size();i++){
25         if(visited[nodes[root].points[i]]==false){    //对子结点进行深搜 
26             dfs(nodes[root].points[i],level+1);
27         }
28     }
29 }
30 
31 int main(){
32     cin>>n;
33     if(n==1){
34         cout<<"1";
35         return 0;
36     }
37     for(int i=1;i<n;i++){
38         int a,b;
39         cin>>a>>b;
40         nodes[a].points.push_back(b);
41         nodes[b].points.push_back(a);
42     }
43     memset(visited,0,sizeof(visited));
44     for(int i=1;i<=n;i++){
45         if(visited[i]==false){
46             visited[i]=true;
47             dfs(i,0);
48             components++;
49         }
50     }
51     if(components>1){
52         cout<<"Error: "<<components<<" components";
53     }else{
54         int k=farthestNode[0];    //距离最远的点
55         farthestNode.clear();
56         height=0;
57         memset(visited,0,sizeof(visited));
58         dfs(k,0);        //从最远点再次进行深搜，得到的点全为deepest root
59         farthestNode.push_back(k);    //k也为deepest root
60         sort(farthestNode.begin(),farthestNode.end());
61         for(int i=0;i<farthestNode.size();i++){
62             cout<<farthestNode[i]<<endl;
63         } 
64     }
65     return 0;
66 }