1021. Deepest Root (25)

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 (<=10000) 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:

51 21 31 42 5

Sample Output 1:

345

Sample Input 2:

51 31 42 53 4

Sample Output 2:

Error: 2 components

具体算法及原理请查阅“树的直径”相关资料，简单叙述一下过程，通过任意一点进行搜索，搜索到的最长点就是树的最深根，然后从任一一个最深根再进行搜索，得到的另外的最深根和第一次的最深根就是这棵树的所有的最深根。也就是连续用两边DFS 或BFS就可以解决问题。这里用了DFS。要注意首先用并查集判断是不是树，然后计算连同分量的个数。

#include <iostream>#include <vector>#include <set>using namespace std;int longest_dist=-1;vector<int> uset;set<int> res;vector<int> visited;vector<vector<int> > tree;int n;void init_uset(){uset.resize(n+1);for(int i=0;i<=n;i++){uset[i]=i;}}int find_root(int x){int r=x;while(uset[r] != r)r=uset[r];int t;while(uset[x] != r){t=uset[x];uset[x]=r;x=t;}return r;}void union_set(int r_x,int r_y){uset[r_x]=r_y;}void dfs(int s,int d,set<int> & ans){bool leaf=true;visited[s]=true;for(int i=0;i<tree[s].size();i++){if(visited[tree[s][i]] == false ){leaf = false;dfs(tree[s][i],d+1,ans);}}if(leaf == true){if(d > longest_dist){longest_dist=d;ans.clear();ans.insert(s);}else if(d == longest_dist){ans.insert(s);}}return;}int main(){cin>>n;tree.resize(n+1);visited.resize(n+1);init_uset();int components=n;for(int i=0;i<n-1;i++){int n1,n2;cin>>n1>>n2;tree[n1].push_back(n2);tree[n2].push_back(n1);int x=find_root(n1);int y=find_root(n2);if(x != y){components--;union_set(x,y);}}if(components != 1){set<int> sets;for(int i=1;i<=n;i++){sets.insert(find_root(i));}printf("Error: %d components\n",(int)sets.size());}else{dfs(1,0,res);longest_dist=-1;set<int> ans;int longest_node=*res.begin();for(int i=1;i<=n;i++)visited[i]=false;dfs(longest_node,0,ans);for(auto itr = res.begin();itr != res.end();++itr)ans.insert(*itr);for(auto itr = ans.begin();itr != ans.end();++itr)cout<<*itr<<endl;}return 0;}

其实不用并查集判断是否是树，用dfs检测连通分量的个数，因为边数为顶点数-1，所以如果连通分量有多个，一定不是树

<pre name="code" class="cpp">#include <iostream>#include <vector>#include <algorithm>#include <set>using namespace std;vector<vector<int> > adjm;vector<int> visited;int deepest=-1;int components=0;int n;set<int> res;set<int> ans;void dfs(int v,int d,set<int> & res){visited[v] = 1;if(d > deepest){deepest = d;res.clear();res.insert(v);}else if(d == deepest){res.insert(v);}for(int i=0;i<adjm[v].size();i++){if(visited[adjm[v][i]] == 0){dfs(adjm[v][i],d+1,res);}}}int main(){cin>>n;adjm.resize(n);visited.resize(n);for(int i=0;i<n-1;i++){int a,b;cin>>a>>b;a--;b--;adjm[a].push_back(b);adjm[b].push_back(a);}for(int i=0;i<n;i++){if(visited[i] == 0){components++;dfs(i,1,res);}}if(components >1){cout<<"Error: "<<components<<" components\n";system("pause");return 0;}for(int i=0;i<n;i++){visited[i] = 0;}deepest = -1;dfs(*res.begin(),1,ans);for(set<int>::iterator itr = res.begin();itr != res.end();++itr)ans.insert(*itr);for(set<int>::iterator itr = ans.begin();itr != ans.end();++itr)cout<<*itr+1<<endl;system("pause");return 0;}