# Educational codeforces round 52f (tree DP, vector)

Source: Internet
Author: User

# Include <bits/stdc ++. h>
Using namespace STD;
Int N, K;
Vector <int> son [1000007];
Int DP [1000007], depth [1000007], ANS [1000007]; // DP [I] indicates the depth of the leaf node closest to I, depth [I] indicates taking I as the root and returning to the number of leaf nodes that I can reach. ans [I] indicates taking I as the root, and the maximum number of leaf nodes can be reached.
Void DFS (INT now ){
If (! Son [now]. Size () {// itself is a leaf node
Depth [now] = 0;
DP [now] = 1;
Return;
}
Int Mn = 1e9, MX = 0;
For (const Int & TMP: Son [now]) {// traverse the child node
DFS (TMP); // continue Deep Search
If (depth [TMP] <k) // if it is less than K, you can go down from now to the leaf node that the child node TMP can reach.
DP [now] + = DP [TMP]; // transmits the leaves that the child can touch to his father.
MX = max (MX, ANS [TMP]-(depth [TMP] <k? DP [TMP]: 0); // depth [TMP] <K, DP [TMP] has been added to DP [now] and removed, MX leaves the largest number of leaf knots that cannot be met once.
Mn = min (Mn, depth [TMP] + 1); // The depth of now is the minimum child depth + 1
}
Depth [now] = Mn; // Mn only places the smallest depth, and none of the subnode's depth is too large, it will only touch the leaf node that meets the meaning of the question once (this time, it will not be able to return to the ancestor node (this DFS parameter)
Ans [now] = DP [now] + mx; // MX can only add one, so it is placed outside the loop.
}
Int main (){
Scanf ("% d", & N, & K );
Int X;
For (INT I = 2; I <= N; I ++ ){
Scanf ("% d", & X );
Son [X]. push_back (I );
}
DFS (1 );
Printf ("% d \ n", ANS [1]);
Return 0;
}

// There is an array to simulate the Linked List's Traversal method.

Educational codeforces round 52f (tree DP, vector)

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