1021 Deepest Root (25分)

问题

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

解决方法

#include
#include
#include
#include
using namespace std;
int n, maxheight = 0, a, b, cnt = 0, s1 = 0;
vector<vector<int>>v;
bool visit[10010];
set<int>s;
vector<int>tmp;
void dfs(int node, int height)
{
	if (height > maxheight)
	{
		tmp.clear(); 
		tmp.push_back(node);
		maxheight = height;
	}
	else if (height == maxheight) tmp.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);
	for (int i = 0; i < n - 1; i++)
	{
		scanf("%d %d", &a, &b);
		v[a].push_back(b);
		v[b].push_back(a);
	}
	for (int i = 1; i <= n; i++)
	{
		if (visit[i] == false)
		{
			dfs(i, 1);
			if (i == 1)
			{
				if (tmp.size() != 0) s1 = tmp[0];
				for (int j = 0; j < tmp.size(); j++) s.insert(tmp[j]);
			}
			cnt++;
		}
	}
	if (cnt >= 2) printf("Error: %d components", cnt);
	else
	{
		tmp.clear();
		maxheight = 0;
		fill(visit, visit + 1010, false);
		dfs(s1, 1);
		for (int i = 0; i < tmp.size(); i++) s.insert(tmp[i]);
		for (auto it = s.begin(); it != s.end(); it++) printf("%d\n", *it);
	}
	return 0;
}