Codeforces Round #670 (Div. 2) C. Link Cut Centroids (dfs traverse tree, centroid)

C. Link Cut Centroids

C. Link Cut Centroids
question meaning: given a tree of n nodes, suppose that among the n nodes, if one of the nodes k (including all the edges contained in the node) is removed, the largest point contained in each connected block The number of is the smallest, then node k is the centroid of the tree.
Now there are two operations

1. Select two nodes $a_1，a_2$And delete the edge between the two
2. Select two nodes $a_1,a_2$Connecting the two points
   requires that the centroid of the tree is unique after two operations.

Idea:
Carefully observe the graph with two centroids given in the title:

Then make a few pictures by hand, you can find

1. Only two centroids at most
2. If there are 2 centroids, then the tree must have symmetry
3. If there are two centroids, there must be no other points between them (because they are symmetrical. If there are an odd number of points in the middle, then the center of mass is the center of mass; if there is an even number of points in the middle, then push to the middle adjacent Point is the centroid)
   then the overall code is relatively simple

• Taking node 1 as the root, dfs traverses the tree, counts the number of sons of each point (including itself), and records it in $son\left[\right]$ array, record the parent nodef [] f[]at the same time$f\left[\right]$ (Two judgments are used in the subsequent counting)
• Traverse all n nodes, assuming that the $The\; i$ nodes are removed, and then each adjacent node is searched for the number of nodes in the connected block where they are:
  • Father node = $n-son[i]$
  • Son node $k$ =$son[k]$
• Just take the maximum value, and then make a structure $cun\left[\right]$ Record the maximum number of nodes in the connected block after removing him$num$ and current own node subscript$id$，$sort$

Finally, look at the num num of the largest and second largest nodesAre$n$$u$$m\; the$ same

• Same: There are two centroids, remove the middle line, and even the son of one to the other
• Not the same: there is only one centroid, just find an edge and add it ^^

int f[maxn], son[maxn], co, n, t;
vector<int> G[maxn];
struct node {
    
    
	int num, id;
}cun[maxn];
bool cmp(node a, node b) {
    
    
	return a.num < b.num;
};
int dfs(int fa, int nw) {
    
    
	f[nw] = fa;
	int sum = 1;
	for (int i = 0; i < G[nw].size(); i++) {
    
    
		if (G[nw][i] == fa)continue;
		sum += dfs(nw, G[nw][i]);
	}
	return son[nw] = sum;
}
int main()
{
    
    
	cin >> t;
	int u, v;
	while (t--)
	{
    
    
		cin >> n;
		for (int i = 1; i <= n; i++)G[i].clear();
		co = 0;
		for (int i = 1; i < n; i++) {
    
    
			cin >> u >> v;
			G[u].push_back(v);
			G[v].push_back(u);
		}
		dfs(0, 1);
		int maxx;
		for (int i = 1; i <= n; i++) {
    
    
			maxx = 0;
			for (int j = 0; j < G[i].size(); j++) {
    
    
				if (G[i][j] == f[i]) {
    
    
					maxx = max(maxx, n - son[i]);
				}
				else {
    
    
					maxx = max(maxx, son[G[i][j]]);
				}
			}
			cun[++co].id = i;
			cun[co].num = maxx;
		}
		sort(cun + 1, cun + 1 + n, cmp);
		int ans;
		if (cun[1].num == cun[2].num) {
    
    
			for (int i = 0; i < G[cun[1].id].size(); i++) {
    
    
				ans = G[cun[1].id][i];
				if (G[cun[1].id][i] != cun[2].id)break;
			}
			cout << cun[1].id << " " << ans << endl;
			cout << ans << " " << cun[2].id << endl;
		}
		else {
    
    
			ans = G[cun[1].id][0];
			cout << cun[1].id << " " << ans << endl;
			cout << cun[1].id << " " << ans << endl;
		}
	}
	return 0;
}