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 ", 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 }