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

  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 }