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hackercupai
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hackercup

#include <algorithm>
#include <cstring>
#include <iostream>
#include <vector>
using namespace std;

const int LIM = 800002;
const int MAXK = 10;

int N, K;
vector<int> children[LIM];

// dp[i][j] = max # of guessable nodes only considering i's subtree,
//            such that i's subtree size is j (with j capping out at K + 1)
int dp[LIM][MAXK + 2];

// dp2[s] = max # of guessable nodes only considering current node's subtree
//          such that its subtree size is s (with s capping out at K + 1)
int dp2[MAXK + 2];

// dp3[a][b] = max # of guessable nodes only considering current node's subtree
//             such that it has a children (with a capping out at K)
//             one of which has subtree size 2 if and only if b = 1,
//             and the rest of which have subtree sizes either 1 or > K
int dp3[MAXK + 1][2];

void rec(int i) {
  // Recursively compute children's DP values.
  for (int c : children[i]) {
    rec(c);
  }
  // Use children's DP values.
  memset(dp2, -1, sizeof dp2);
  memset(dp3, -1, sizeof dp3);
  dp2[1] = dp3[0][0] = 0;
  for (int c : children[i]) {
    // Assume no new guessable nodes.
    for (int s1 = K + 1; s1 >= 1; s1--) {
      int d1 = dp2[s1];
      if (d1 >= 0) {
        for (int s2 = 1; s2 <= K + 1; s2++) {
          int d2 = dp[c][s2];
          if (d2 >= 0) {
            int s = min(K + 1, s1 + s2);
            dp2[s] = max(dp2[s], d1 + d2);
          }
        }
      }
    }
    // New guessable node, situation 1:
    // - Node i has 1 child
    // - That child's subtree size is > K (that child is guessable)
    int d1 = dp[c][K + 1];
    if (d1 >= 0) {
      dp[i][K + 1] = max(dp[i][K + 1], d1 + 1);
    }
    // New guessable node, situation 2:
    // - Node i has at least K children
    // - One child's subtree size is exactly 2 (that child's single child is
    // guessable)
    // - Each other child's subtree size is either exactly 1, or > K
    for (int a1 = K; a1 >= 0; a1--) {
      for (int b1 : {1, 0}) {
        d1 = dp3[a1][b1];
        if (d1 >= 0) {
          int a2 = min(K, a1 + 1);
          int d2 = max(dp[c][1], dp[c][K + 1]);
          if (d2 >= 0) {
            dp3[a2][b1] = max(dp3[a2][b1], d1 + d2);
          }
          if (!b1 && (d2 = dp[c][2]) >= 0) {
            dp3[a2][1] = max(dp3[a2][1], d1 + d2);
          }
        }
      }
    }
  }
  // Merge sub-DP results into main DP.
  for (int s1 = 1; s1 <= K + 1; s1++) {  // No new guessable nodes.
    dp[i][s1] = max(dp[i][s1], dp2[s1]);
  }
  int d1 = dp3[K][1];
  if (d1 >= 0) {  // New guessable node, situation 2.
    dp[i][K + 1] = max(dp[i][K + 1], d1 + 1);
  }
}

int solve() {
  for (int i = 0; i < N; i++) {
    children[i].clear();
  }
  // Input.
  cin >> N >> K;
  for (int i = 1; i < N; i++) {
    int p;
    cin >> p;
    children[p - 1].push_back(i);
  }
  // Tree DP.
  memset(dp, -1, sizeof(dp[0]) * N);
  rec(0);
  // Compute answer.
  int ans = 0;
  for (int i = 1; i <= K + 1; i++) {
    ans = max(ans, dp[0][i]);
  }
  return ans;
}

int main() {
  int T;
  cin >> T;
  for (int t = 1; t <= T; t++) {
    cout << "Case #" << t << ": " << solve() << endl;
  }
  return 0;
}

	#include <algorithm>
	#include <cstring>
	#include <iostream>
	#include <vector>
	using namespace std;

	const int LIM = 800002;
	const int MAXK = 10;

	int N, K;
	vector<int> children[LIM];

	// dp[i][j] = max # of guessable nodes only considering i's subtree,
	// such that i's subtree size is j (with j capping out at K + 1)
	int dp[LIM][MAXK + 2];

	// dp2[s] = max # of guessable nodes only considering current node's subtree
	// such that its subtree size is s (with s capping out at K + 1)
	int dp2[MAXK + 2];

	// dp3[a][b] = max # of guessable nodes only considering current node's subtree
	// such that it has a children (with a capping out at K)
	// one of which has subtree size 2 if and only if b = 1,
	// and the rest of which have subtree sizes either 1 or > K
	int dp3[MAXK + 1][2];

	void rec(int i) {
	// Recursively compute children's DP values.
	for (int c : children[i]) {
	rec(c);
	}
	// Use children's DP values.
	memset(dp2, -1, sizeof dp2);
	memset(dp3, -1, sizeof dp3);
	dp2[1] = dp3[0][0] = 0;
	for (int c : children[i]) {
	// Assume no new guessable nodes.
	for (int s1 = K + 1; s1 >= 1; s1--) {
	int d1 = dp2[s1];
	if (d1 >= 0) {
	for (int s2 = 1; s2 <= K + 1; s2++) {
	int d2 = dp[c][s2];
	if (d2 >= 0) {
	int s = min(K + 1, s1 + s2);
	dp2[s] = max(dp2[s], d1 + d2);
	}
	}
	}
	}
	// New guessable node, situation 1:
	// - Node i has 1 child
	// - That child's subtree size is > K (that child is guessable)
	int d1 = dp[c][K + 1];
	if (d1 >= 0) {
	dp[i][K + 1] = max(dp[i][K + 1], d1 + 1);
	}
	// New guessable node, situation 2:
	// - Node i has at least K children
	// - One child's subtree size is exactly 2 (that child's single child is
	// guessable)
	// - Each other child's subtree size is either exactly 1, or > K
	for (int a1 = K; a1 >= 0; a1--) {
	for (int b1 : {1, 0}) {
	d1 = dp3[a1][b1];
	if (d1 >= 0) {
	int a2 = min(K, a1 + 1);
	int d2 = max(dp[c][1], dp[c][K + 1]);
	if (d2 >= 0) {
	dp3[a2][b1] = max(dp3[a2][b1], d1 + d2);
	}
	if (!b1 && (d2 = dp[c][2]) >= 0) {
	dp3[a2][1] = max(dp3[a2][1], d1 + d2);
	}
	}
	}
	}
	}
	// Merge sub-DP results into main DP.
	for (int s1 = 1; s1 <= K + 1; s1++) { // No new guessable nodes.
	dp[i][s1] = max(dp[i][s1], dp2[s1]);
	}
	int d1 = dp3[K][1];
	if (d1 >= 0) { // New guessable node, situation 2.
	dp[i][K + 1] = max(dp[i][K + 1], d1 + 1);
	}
	}

	int solve() {
	for (int i = 0; i < N; i++) {
	children[i].clear();
	}
	// Input.
	cin >> N >> K;
	for (int i = 1; i < N; i++) {
	int p;
	cin >> p;
	children[p - 1].push_back(i);
	}
	// Tree DP.
	memset(dp, -1, sizeof(dp[0]) * N);
	rec(0);
	// Compute answer.
	int ans = 0;
	for (int i = 1; i <= K + 1; i++) {
	ans = max(ans, dp[0][i]);
	}
	return ans;
	}

	int main() {
	int T;
	cin >> T;
	for (int t = 1; t <= T; t++) {
	cout << "Case #" << t << ": " << solve() << endl;
	}
	return 0;
	}