【PAT】A-1021 Deepest Root【树】【树的深度】【DFS】

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

题目大意

给出n个结点（1~n）之间的n条边，问是否能构成一棵树，如果不能构成则输出它有的连通分量个数，如果能构成一棵树，输出能构成最深的树的高度时，树的根结点。如果有多个，按照从小到大输出。

思路

进行两次深度优先搜索，第一次任取一个结点作为根结点（为了方便起见选择1号作为根结点），将这一次遍历中最深的结点（可能不止一个）加入到向量depthestNodes中，同时计算连通分量。

• 若连通分量大于1，则输出Error: x components。
• 否则执行第二次深度优先搜索，根结点为上一轮depthestNodes中的任意一个（为了方便取第一个）。

代码

#include <iostream>
#include <cstdio>
#include <vector>
#include <set>
#define maxN 10001
using namespace std;
bool marked[maxN];
vector<int> graph[maxN];
int maxHeight = 0;
vector<int> depthestNodes;