PAT - 甲级 - 1021. Deepest Root (25)（DFS图的连通分量，最大深度）

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

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给定条件：
1.n个节点
2.n-1条连线

题目要求：
1.判断给定的图是否连通的
2.如果连通，求出选取那些节点作为根节点，所建成的树的深度最大
3.按照编号从小到大的顺序，输出符合条件的根节点

解决方法：
1.dfs求图的连通分量，判断是否为1
2.分别选取所有的节点作为根节点，进行深度优先遍历，记录所到达的最大深度，并且在搜索过程中记录每个节点所能处于的最大深度
3.遍历所有节点，输出所在深度可以为最大深度的节点

---

#include<cstdio>
#include<vector>
using namespace std;

int N, a, b, maxd = -1;
int vis[10001];
int depth[10001];
vector<int> e[10001];

void dfs(int v, int d){
	vis[v] = 1;
	depth[v] = max(depth[v], d); // update node's depth
	maxd = max(maxd, d); // update the max depth
	for(int i = 0; i < e[v].size(); i++){
		int next = e[v][i];
		if(!vis[next]){
			dfs(next,d+1);
		}
	}
}

int main(){
	while(scanf("%d", &N) != EOF){
		// input
		for(int i = 1; i < N; i++){
			scanf("%d%d", &a, &b);
			e[a].push_back(b);
			e[b].push_back(a);
		}
		fill(vis, vis+10001, 0);

		int cnt = 0;
		for(int i = 1; i <= N; i++){
			if(!vis[i]){
				dfs(i,1);
				cnt++; // the number of components
			}	
		}
		if(cnt != 1){
			printf("Error: %d components\n",cnt);
		}else{
			// dfs every node
			for(int i = 1; i <= N; i++){
				fill(vis, vis+10001, 0);
				vis[i] = 1;
				dfs(i,1);
			}
			// find the node which depth is maxd
			for(int i = 1; i <=N; i++){
				if(depth[i] == maxd){
					printf("%d\n", i);
				}
			}
		}
	}
	return 0;
}