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 (≤104) 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
#include<iostream>
#include<vector>
#include<set>
using namespace std;
int n;
int maxheight = 0;
vector<vector<int>> v;
bool visit[10010];
vector<int> temp;
set<int> s;
void dfs(int node,int height){
if(height>maxheight){
temp.clear();
temp.push_back(node);
maxheight = height;
}else if(height==maxheight){
temp.push_back(node);
}
visit[node] = true;
for(int i=0;i<v[node].size();i++){
if(visit[v[node][i]]==false){
dfs(v[node][i],height+1);
}
}
}
int main(){
scanf("%d",&n);
v.resize(n+1);
int a,b;
for(int i=0;i<n-1;++i){
scanf("%d%d",&a,&b);
v[a].push_back(b);
v[b].push_back(a);
}
int cnt = 0;
int e;
for(int i=1;i<=n;i++){//节点是以1开始
if(visit[i]==false){
dfs(i,1);
if(i==1){
if(temp.size()!=0){
//随意取一个使树深度最大的节点
//两次dfs求并
e = temp[0];
}
for(int j=0;j<temp.size();++j){
s.insert(temp[j]);
}
}
cnt++;
}
}
if(cnt>=2){
printf("Error: %d components",cnt);
}else{
temp.clear();
maxheight = 0;
fill(visit,visit+10010,false);
dfs(e,1);
for(int j=0;j<temp.size();++j){
s.insert(temp[j]);
}
for(auto it=s.begin();it!=s.end();++it){
printf("%d\n",*it);
}
}
return 0;
}
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