POI #Year2006 #妙妙题 #贪心

考虑从下往上按照拓扑序分层，对于每一层，这一层最多可以选择 \(min(2m,cnt)\) 个

考虑这个上界是否可以达到，这是一定可以的，通过将在下面结束的路径向上，可以做到每个点都被经过

所以直接统计

// Author: xiaruize
#ifndef ONLINE_JUDGE
#define debug(x) cerr << "On Line:" << __LINE__ << #x << "=" << x << endl
bool start_of_memory_use;
#else
#define debug(x)
#endif
#include <bits/stdc++.h>
using namespace std;
#ifndef ONLINE_JUDGE
clock_t start_clock = clock();
#endif
// #define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define mms(arr, n) memset(arr, n, sizeof(arr))
#define rep(i, a, n) for (int i = (a); i <= (n); ++i)
#define per(i, n, a) for (int i = (n); i >= (a); --i)
int max(int a, int b)

{
	if (a > b)
		return a;
	return b;
}
int min(int a, int b)
{
	if (a < b)
		return a;
	return b;
}
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 1e6 + 10;

int n, m;
vector<int> g[N];
int cnt[N], dep[N];
int deg[N];

void solve()
{
	cin >> n >> m;
	rep(i, 1, n - 1)
	{
		int u, v;
		cin >> u >> v;
		g[u].push_back(v);
		g[v].push_back(u);
		deg[u]++;
		deg[v]++;
	}
	queue<int> q;
	rep(i, 1, n)
	{
		if (deg[i] == 1)
		{
			dep[i] = 1;
			q.push(i);
		}
	}
	while (!q.empty())
	{
		int x = q.front();
		q.pop();
		cnt[dep[x]]++;
		for (auto v : g[x])
		{
			dep[v] = dep[x] + 1;
			deg[v]--;
			if (deg[v] == 1)
				q.push(v);
		}
	}
	int res = 0;
	rep(i, 1, n)
	{
		res += min(cnt[i], m * 2);
		// cerr << cnt[i] << ' ';
	}
	cout << res << endl;
}

#ifndef ONLINE_JUDGE
bool end_of_memory_use;
#endif

signed main()
{
	// freopen(".in","r",stdin);
	// freopen(".out","w",stdout);
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int testcase = 1;
	// cin >> testcase;
	while (testcase--)
		solve();
#ifndef ONLINE_JUDGE
	cerr << "Memory use:" << (&end_of_memory_use - &start_of_memory_use) / 1024.0 / 1024.0 << "MiB" << endl;
	cerr << "Time use:" << (double)clock() / CLOCKS_PER_SEC * 1000.0 << "ms" << endl;
#endif
	return 0;
}