PAT甲级 1021 Deepest Root

PAT甲级 1021 Deepest Root

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 $N (≤10^4)$ 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

这道题一开始没啥思路，想暴力然后T了，后来看了图论的知识懂了，要找符合条件的根节点：
1.在树上任取一点 $node1$ ， $DFS$ 得到最深结点集合 $A$
2.在从集合 $A$ 任取一点 $node2$ ， $DFS$ 得到最深结点集合 $B$
3.答案即为 $A∪B$
$PS$：注意特判 $n=1$ 的情况
代码如下：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e4+5;
vector<int>G[N];
set<int>s;//A∪B,选set可以去重且自动排序
int n,u,v,cnt;
int deep[N],vis[N];
int ans=0,maxdeep=1;
void dfs(int u){
    vis[u]=1;
    for(int v:G[u]){
        if(!vis[v]){
            deep[v]=deep[u]+1;
            maxdeep=max(maxdeep,deep[v]);
            dfs(v);
        }
    }
}

int main() {
	scanf("%d",&n);
	for(int i=0;i<n-1;i++){
        scanf("%d%d",&u,&v);
        G[u].push_back(v);
        G[v].push_back(u);
	}
	cnt=0;
	memset(vis,0,sizeof(vis));
	deep[1]=1;maxdeep=0;
	int p;//记录A中任一最深结点
	for(int i=1;i<=n;i++){
        if(!vis[i]) {dfs(i);cnt++;}//第一遍DFS得到联通块的数量
        if(deep[i]==maxdeep){
            s.insert(i);p=i;//集合A
        }
	}
	memset(vis,0,sizeof(vis));
	deep[p]=1;maxdeep=0;
	dfs(p);
	for(int i=1;i<=n;i++){
        if(deep[i]==maxdeep){
            s.insert(i);//集合B
        }
	}
	if(n==1) puts("1");
	else if(cnt>1){
        printf("Error: %d components",cnt);
	}
	else{
        for(set<int>::iterator i=s.begin();i!=s.end();i++)
            printf("%d\n",*i);
	}
	return 0;
}