The nation of Panel holds an annual show called The Number Games, where each district in the nation will be represented by one contestant.

The nation has $\mathrm{nn\; districts\; numbered\; from}$

This year, the president decided to reduce the costs. He wants to remove $\mathrm{kk\; contestants\; from\; the\; games.\; However,\; the\; districts\; of\; the\; removed\; contestants\; will\; be\; furious\; and\; will\; not\; allow\; anyone\; to\; cross\; through\; their\; districts.}$

The president wants to ensure that all remaining contestants are from districts that can be reached from one another. He also wishes to maximize the total number of fans of the participating contestants.

Which contestants should the president remove?

The first line of input contains two integers

The next $\mathrm{n?1n?1\; lines\; each\; contains\; two\; integers\; aa\; and\; bb\; (1\le a,b\le n1\le a,b\le n,\; a\ne ba\ne b),\; that\; describe\; a\; road\; that\; connects\; two\; different\; districts\; aaand\; bb\; in\; the\; nation.\; It\; is\; guaranteed\; that\; there\; is\; exactly\; one\; path\; between\; every\; two\; districts.}$

Print $\mathrm{kk\; space-separated\; integers:\; the\; numbers\; of\; the\; districts\; of\; which\; the\; contestants\; should\; be\; removed,\; in\; increasing\; order\; of\; district\; number.}$

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

1 3 4

8 4
2 6
2 7
7 8
1 2
3 1
2 4
7 5

1 3 4 5

In the first sample, the maximum possible total number of fans is ${\mathrm{22+25+26=10022+25+26=100.\; We\; can\; achieve\; it\; by\; removing\; the\; contestants\; of\; the\; districts\; 1,\; 3,\; and\; 4.}}^{}$

  1 #include <iostream>
  2 #include <fstream>
  3 #include <sstream>
  4 #include <cstdlib>
  5 #include <cstdio>
  6 #include <cmath>
  7 #include <string>
  8 #include <cstring>
  9 #include <algorithm>
 10 #include <queue>
 11 #include <stack>
 12 #include <vector>
 13 #include <set>
 14 #include <map>
 15 #include <list>
 16 #include <iomanip>
 17 #include <cctype>
 18 #include <cassert>
 19 #include <bitset>
 20 #include <ctime>
 21 
 22 using namespace std;
 23 
 24 #define pau system("pause")
 25 #define ll long long
 26 #define pii pair<int, int>
 27 #define pb push_back
 28 #define mp make_pair
 29 #define clr(a, x) memset(a, x, sizeof(a))
 30 
 31 const double pi = acos(-1.0);
 32 const int INF = 0x3f3f3f3f;
 33 const int MOD = 1e9 + 7;
 34 const double EPS = 1e-9;
 35 
 36 /*
 37 #include <ext/pb_ds/assoc_container.hpp>
 38 #include <ext/pb_ds/tree_policy.hpp>
 39 
 40 using namespace __gnu_pbds;
 41 tree<pli, null_type, greater<pli>, rb_tree_tag, tree_order_statistics_node_update> T;
 42 */
 43 
 44 int n, k, d[1000015], vis[1000015], p[1000015];
 45 vector<int> e[1000015], e1[1000015];
 46 int f[1000015][25];
 47 
 48 void dfs(int x) {
 49     vis[x] = 1;
 50     for (int i = 0; i < e[x].size(); ++i) {
 51         int y = e[x][i];
 52         if (vis[y]) p[x] = y;
 53         else {
 54             e1[x].pb(y);
 55             dfs(y);
 56         }
 57     }
 58 }
 59 
 60 void dfs2(int x) {
 61     f[x][0] = p[x];
 62     for (int i = 1; i <= 20; ++i) {
 63         f[x][i] = f[f[x][i - 1]][i - 1];
 64     }
 65     for (int i = 0; i < e1[x].size(); ++i) {
 66         int y = e1[x][i];
 67         dfs2(y);
 68     }
 69 }
 70 vector<int> ans;
 71 bool out[1000015];
 72 void solve() {
 73     clr(vis, 0); vis[0] = 1;
 74     int now = n;
 75     while (k) {
 76         if (vis[now]) {
 77             --now;
 78             continue;
 79         }
 80         int cnt = 1;
 81         for (int h = 0, i = now; ; ) {
 82             while (!vis[f[i][h]]) {
 83                 ++h;
 84             }
 85             while (~h && vis[f[i][h]]) {
 86                 --h;
 87             }
 88             if (~h) {
 89                 cnt += 1 << h;
 90                 i = f[i][h];
 91             } else {
 92                 break;
 93             }
 94         }
 95         if (cnt <= k) {
 96             int i = now;
 97             while (!vis[i]) {
 98                 ans.pb(i);
 99                 vis[i] = 1;
100                 i = p[i];
101             }
102             k -= cnt;
103         }
104         --now;
105     }
106     for (int i = 0; i < ans.size(); ++i) {
107         out[ans[i]] = 1;
108     }
109     for (int i = 1; i <= n; ++i) {
110         if (!out[i]) printf("%d ", i);
111     }
112 }
113 int main() {
114     scanf("%d%d", &n, &k);
115     k = n - k;
116     for (int i = 1; i < n; ++i) {
117         int a, b;
118         scanf("%d%d", &a, &b);
119         e[a].pb(b), e[b].pb(a);
120     }
121     dfs(n);
122     dfs2(n);
123     solve();
124     return 0;
125 }

Codeforces Round #480 (Div. 2) E. The Number Games