Problem:
Score : 600 points

Problem Statement

Given an undirected tree, let the distance between vertices u and v be the
number of edges on the simple path from u to v.
The diameter of a tree is the maximum among the distances between any two vertices.
We will call a tree good if and only if its diameter is at most K.

You are given an undirected tree with N vertices numbered 1 through N.
For each i (1≦i≦N-1), there is an edge connecting vertices A_i and B_i.

You want to remove zero or more vertices from the tree, so that the resulting tree is good.
When a vertex is removed, all incident edges will also be removed.
The resulting graph must be connected.

Find the minimum number of vertices that you need to remove in order to produce a good tree.

Constraints

2≦N≦2000

1≦K≦N-1

1≦A_i≦N, 1≦B_i≦N

The graph defined by A_i and B_i is a tree.

Input

The input is given from Standard Input in the following format:

N K
A_1 B_1
A_2 B_2
:
A_{N-1} B_{N-1}

Output

Print the minimum number of vertices that you need to remove in order to produce a good tree.

Sample Input 1

6 2
1 2
3 2
4 2
1 6
5 6

Sample Output 1

2

The tree is shown below. Removing vertices 5 and 6 will result in a good tree with the diameter of 2.

Sample Input 2

6 5
1 2
3 2
4 2
1 6
5 6

Sample Output 2

0

Since the given tree is already good, you do not need to remove any vertex.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <random>

using namespace std;

typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;
//#define int ll
//#define int __int128
//#define ll __int128
typedef pair<int, int   > pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<pii> vpi;
typedef vector< vi > vvi;
typedef vector< vvi > vvvi;
typedef vector<short> vs;
typedef vector<vs> vvs;
typedef vector<vvs> vvvs;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<vvl> vvvl;
typedef vector<ld> vld;
typedef vector<vld> vvld;
typedef vector<vvld> vvvld;
typedef vector<string> vst;
typedef vector<vst> vvst;
typedef pair<ld, ld> pld;
//typedef complex<double> base;

#define inmin(a, b) a = min(a, (b))
#define inmax(a, b) a = max(a, (b))
#define ALL(a) a.begin(),a.end()
#define RALL(a) a.rbegin(),a.rend()
#define sqr(x) ((x) * (x))
#define fori(i, n) for(int i = 0; i < (int)(n); ++i)
#define SZ(a) ((int)((a).size()))
#define triple(T) tuple<T, T, T>
#define quad(T) tuple<T, T, T, T>
#define watch(x) cerr << (#x) << " = " << (x) << endl;
#define mp(a, b) make_pair(a, b)

#ifdef MAX_HOME
#define cerr cout
#else
#define cerr if (false) cerr
#endif

const double PI = 2 * acos(0.0);
#define rand shittttty_shit
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng_64(chrono::steady_clock::now().time_since_epoch().count());

const string DIGITS = "0123456789";
const string ALPH = "abcdefghijklmnopqrstuvwxyz";

template <class T0, class T1>
inline ostream & operator << (ostream &out, pair<T0, T1> &a) {
    return out << "{" << a.first << ", " << a.second << "}";
}

template <class T0, class T1, class T2>
inline ostream & operator << (ostream &out, tuple<T0, T1, T2> &a) {
    return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a) << "}";
}

template <class T0, class T1, class T2, class T3>
inline ostream & operator << (ostream &out, tuple<T0, T1, T2, T3> &a) {
    return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a) << ", " <<  get<3>(a) << "}";
}

template<class T>
inline ostream & operator << (ostream &out, vector<T> &a) {
    out << "[";
    fori (i, a.size())
        out << a[i] << vector<string>{", ", "]  "}[i + 1 == a.size()];
    return out;
}



void smain();

signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

#ifdef MAX_HOME
    freopen("input.txt", "r", stdin);
    clock_t start = clock();
#endif
    cout << setprecision(12) << fixed;
    smain();
#ifdef MAX_HOME
    cout << "\n\n\n\nTOTAL EXECUTION TIME: " << float( clock () - start ) /  CLOCKS_PER_SEC << endl;
#endif
    return 0;
}

const int N = 2020;
int n, k;

vi g[N];

void dfs(int v, int par, vi & sack, int d = 1) {
    while (sack.size() <= d)
        sack.push_back(0);
    sack[d]++;
    for (int to : g[v]) {
        if (to == par)
            continue;
        dfs(to, v, sack, d + 1);
    }
}

void smain() {
    cin >> n >> k;
    fori (i, n - 1) {
        int u, v;
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }

    int ans = 0;

    for (int v = 1; v <= n; ++v) {
        vvi sacks;
        for (int to : g[v]) {
            sacks.push_back({});
            dfs(to, v, sacks.back());
        }
        if (k & 1) {
            int sum = 1;
            int h = k >> 1;
            for (auto sack : sacks) {
                for (int i = 0; i < min((int) sack.size(), h + 1); ++i) {
                    sum += sack[i];
                }
            }
            for (auto sack : sacks) {
                if (sack.size() <= h + 1)
                    continue;
                int cur = sum + sack[h + 1];
                inmax(ans, cur);
            }
        } else {
            int h = k >> 1;
            int cur = 1;
            for (auto sack : sacks) {
                for (int i = 0; i < min((int) sack.size(), h + 1); ++i) {
                    cur += sack[i];
                }
            }
            inmax(ans, cur);
        }
    }
    cout << n - ans;
}