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luogu P2014 AC

Jan 23rd, 2023
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  1. // Problem: P2014 [CTSC1997] 选课
  2. // Contest: Luogu
  3. // URL: https://www.luogu.com.cn/problem/P2014
  4. // Memory Limit: 125 MB
  5. // Time Limit: 1000 ms
  6. //
  7. // Powered by CP Editor (https://cpeditor.org)
  8.  
  9. #include <assert.h>
  10. #include <bits/stdc++.h>
  11. using namespace std;
  12. #define dbg(...) logger(#__VA_ARGS__, __VA_ARGS__)
  13. template <typename... Args> void logger(string vars, Args &&... values)
  14. {
  15.     cerr << vars << " = ";
  16.     string delim = "";
  17.     (..., (cerr << delim << values, delim = ", "));
  18.     cerr << endl;
  19. }
  20.  
  21. template <class T> inline auto vv(int m) { return vector<vector<T>>(m, vector<T>(m)); }
  22. template <class T> inline auto vv(int m, int n) { return vector<vector<T>>(m, vector<T>(n)); }
  23. template <class T, T init> inline auto vv(int m) { return vector<vector<T>>(m, vector<T>(m, init)); }
  24. template <class T, T init> inline auto vv(int m, int n) { return vector<vector<T>>(m, vector<T>(n, init)); }
  25.  
  26. template <class T> using mpq = priority_queue<T, vector<T>, greater<T>>;
  27.  
  28. using ll = long long;
  29. using pii = pair<int, int>;
  30. using vl = vector<ll>;
  31. using vi = vector<int>;
  32.  
  33. vi g[512];
  34. vi score;
  35. int n, m;
  36. vector<vector<int>> dp;
  37.  
  38. int dfs(int u)
  39. {
  40.     int cnt = 1;
  41.     dp[u][0] = 0;
  42.     dp[u][1] = score[u];
  43.     for (auto v : g[u]) {
  44.         auto subcnt = dfs(v);
  45.         cnt += subcnt;
  46.         for (int j = cnt; j > 0; --j)
  47.             for (int k = 1; k <= subcnt; ++k)
  48.                 if (j - k > 0)
  49.                     dp[u][j] = max(dp[u][j], dp[u][j - k] + dp[v][k]);
  50.     }
  51.     return cnt;
  52. }
  53.  
  54. int main(int argc, char **argv)
  55. {
  56.     cin >> n >> m;
  57.     score = vi(n + 1);
  58.     for (int i = 1; i <= n; ++i) {
  59.         int k;
  60.         cin >> k >> score[i];
  61.         g[k].push_back(i);
  62.     }
  63.     dp = vv<int, INT_MIN / 2>(n + 1, n + 1);
  64.     dfs(0);
  65.     cout << dp[0][m + 1] << endl;
  66.     return 0;
  67. };
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