1021 Deepest Root (25 分)

1021 Deepest Root (25 分)

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

终于不是模拟题了，舒服。。。大概题意就是----给你一组数据，要么是树，要么就是非连通图，如果是树，你就把最远距离的端点输出来，如果不是树，就输出它分成了几个部分。

题目挺好的。题解如下：26ms

 1 #include <bits/stdc++.h>
 2 #define N 10050
 3 using namespace std;
 4 int n, pos = 0;
 5 vector<int> v[N];
 6 int vis[N], val[N];
 7 int max_len = 0, id = 0;
 8 set<int> s;
 9 set<int>::iterator it;
10 bool flag = false;
11 void dfs(int x){
12     vis[x] = 1;
13     for(int i = 0 ; i < v[x].size(); i++){
14         int k = v[x][i];
15         if(vis[k] == 0)
16             dfs(k);
17     }
18 }
19 void dfs1(int x,int len){
20     vis[x] = 1;
21     if(len > max_len){
22         max_len = len;
23         id = x;
24     }
25     for(int i = 0; i < v[x].size(); i++){
26         int k = v[x][i];
27         if(vis[k] == 0){
28             dfs1(k, len+1);
29         }
30     }
31 }
32
33 void dfs2(int x,int len){
34     vis[x] = 1;
35     if(len == max_len){
36         s.insert(x);
37         val[x] = 1;
38         flag = true;
39     }
40     for(int i = 0; i < v[x].size(); i++){
41         int k = v[x][i];
42         if(vis[k] == 0){
43             dfs2(k, len+1);
44         }
45     }
46 }
47
48 int main(){
49     cin >> n;
50     if(n == 1){
51         cout << "1" << endl;
52         return 0;
53     }
54     int x, y;
55     for(int i = 1; i < n; i++){
56         cin >> x >> y;
57         v[x].push_back(y);
58         v[y].push_back(x);
59     }
60     memset(vis,0,sizeof(vis));
61     for(int i = 1; i <= n; i++){
62         if(vis[i] == 0){
63             dfs(i);
64             pos++;
65         }
66     }
67     if(pos > 1){
68         printf("Error: %d components\n", pos);
69     }else{
70         memset(vis,0,sizeof(vis));
71         dfs1(1,0);
72         memset(vis,0,sizeof(vis));
73         dfs1(id,0);
74         for(int i = 1; i <= n; i++){
75             if(v[i].size() == 1 && val[i] == 0){
76                 memset(vis,0,sizeof(vis));
77                 dfs2(i,0);
78                 if(flag){
79                     flag = false;
80                     s.insert(i);
81                     val[i] = 1;
82                 }
83             }
84         }
85         // cout << "****" <<endl;
86         for(it = s.begin(); it != s.end(); it++)
87             cout << *it << endl;
88     }
89     return 0;
90 }

原文地址：https://www.cnblogs.com/zllwxm123/p/11048701.html