PAT (Advanced Level) Practise 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:

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

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;
}