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 1:5 1 2 1 3 1 4 2 5Sample Output 1:
3 4 5Sample Input 2:
5 1 3 1 4 2 5 3 4Sample Output 2:
Error: 2 components
代码:
#include <cstdio> #include <cstring> #include <vector> #include <algorithm> using namespace std; const int maxn=10010; vector<int> G[maxn]; int tree[maxn]; int findRoot(int x){ if(tree[x]==-1) return x; else{ int tmp=findRoot(tree[x]); tree[x]=tmp; return tmp; } } int maxh=0; vector<int> temp,ans; void dfs(int u,int h,int pre){ if(h>maxh){ temp.clear(); temp.push_back(u); maxh=h; } else if(h==maxh) temp.push_back(u); for(int i=0;i<G[u].size();i++){ if(G[u][i]==pre) continue; dfs(G[u][i],h+1,u); } } int main(){ int n,a,b; scanf("%d",&n); for(int i=1;i<=n;i++) tree[i]=-1; for(int i=1;i<n;i++){ scanf("%d%d",&a,&b); G[a].push_back(b); G[b].push_back(a); a=findRoot(a); b=findRoot(b); if(a!=b) tree[a]=b; } int block=0; for(int i=1;i<=n;i++) if(tree[i]==-1) block++; if(block!=1) printf("Error: %d components\n",block); else{ dfs(1,1,-1); ans=temp; dfs(ans[0],1,-1); for(int i=0;i<temp.size();i++) ans.push_back(temp[i]); sort(ans.begin(),ans.end()); printf("%d\n",ans[0]); for(int i=1;i<ans.size();i++){ if(ans[i]!=ans[i-1]) printf("%d\n",ans[i]); } } return 0; }