1021 Deepest Root (25)

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

解题思路：
图的dfs，计算连通分量个数。
先做一次dfs，如果图连通，即dfs次数，若不为1，输出Error: x components，若为1，从第一次dfs得到的最深节点开始，再次dfs，两次得到的节点集合（并集）就是所求最深节点。
这题自己难以想到的是两次dfs得到最深节点。

#include<cstdio>
#include<set>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn=1e5+10;
int n,deepest=0;
vector<int>adj[maxn],temp;
bool vis[maxn]={false};
set<int>s;

void dfs(int v,int deep){
    vis[v]=true;
    if(deep>deepest){
        deepest=deep;
        temp.clear();
        temp.push_back(v) ;
    }else if(deep==deepest){
        temp.push_back(v)  ;
    }
    for(int i=0;i<adj[v].size() ;i++){
        if(vis[adj[v][i]]==false)
            dfs(adj[v][i],deep+1);
    }
}
int main(){
    int a,b,cnt=0;
    scanf("%d",&n);
    for(int i=0;i<n-1;i++){
        scanf("%d%d",&a,&b);
        adj[a].push_back(b);
        adj[b].push_back(a);
    }
    fill(vis,vis+maxn,false);
    for(int i=1;i<=n;i++){
        if(vis[i]==false){
            dfs(i,1);
            cnt++;
        }
    }
    if(cnt!=1){
        printf("Error: %d components",cnt);
    }else{
        int s1=temp[0];
        for(int i=0;i<temp.size() ;i++)
            s.insert(temp[i]) ;
        temp.clear() ;
        deepest=0;
        fill(vis,vis+maxn,false);
        dfs(s1,1);
        for(int i=0;i<temp.size() ;i++) 
            s.insert(temp[i]) ;
        for(set<int>::iterator it=s.begin() ;it!=s.end() ;it++){
            printf("%d\n",*it);
        }
    }
    return 0;
}