Cow Pedigrees

This is a explanation of this problem from USACO's training website. I have converted it to markdown. Please do not just copy code; you will not learn anything; at least type it out and understand so you can do it yourself in the future!

Farmer John is considering purchasing a new herd of cows. In this new herd, each mother cow gives birth to two children. The relationships among the cows can easily be represented by one or more binary trees with a total of N (3 <= N < 200) nodes. The trees have these properties:

The degree of each node is 0 or 2. The degree is the count of the node’s immediate children.
The height of the tree is equal to K (1 < K < 100). The height is the number of nodes on the longest path from the root to any leaf; a leaf is a node with no children.

How many different possible pedigree structures are there? A pedigree is different if its tree structure differs from that of another pedigree. Output the remainder when the total number of different possible pedigrees is divided by 9901.

PROGRAM NAME: nocows

INPUT FORMAT

Line 1: Two space-separated integers, N and K.

SAMPLE INPUT (file nocows.in)

5 3

OUTPUT FORMAT

Line 1: One single integer number representing the number of possible pedigrees MODULO 9901.

SAMPLE OUTPUT (file nocows.out)

OUTPUT DETAILS

Two possible pedigrees have 5 nodes and height equal to 3:

    @                   @      
   / \                 / \
  @   @      and      @   @
 / \                     / \
@   @                   @   @

CODE

Java

C++

Pascal

ANALYSIS

Silviu Ganceanu

This is a DP problem. The properties of a tree that we are interested in are depth and number of nodes, so we’ll make a table: table[i][j] contains the number of trees with depth i and number of nodes j. Given the constraints of the task, j must be odd. How do you construct a tree? From smaller trees, of course. A tree of depth i and j nodes will be constructed from two smaller trees and one more node.

With i and j already chosen, we chose k, which is the number of nodes in the left subtree. Then the number of nodes in the right subtree is known, j-k-1. For depth, at least one subtree has to have depth i-1 so that the new made tree would have depth i. There are three possibilities: the left subtree can have depth i-1 and the depth of the right subtree can be smaller, the right subtree can have depth i-1 and the depth of the left subtree can be smaller, or they can both have depth i-1.

The truth is that once we are constructing trees of depth i, we use smaller trees, but we only care if those are of depth i-1 or smaller. So, let another array, smalltrees[i-2][j] contain number of trees of any depth smaller than i-1, not just i-2. Now, knowing all this, we contruct our tree from three possible ways:

table[i][j] += smalltrees[i-2][k]*table[i-1][j-1-k];
                  // left subtree smaller than i-1, right is i-1
table[i][j] += table[i-1][k]*smalltrees[i-2][j-1-k];
                  // left subtree is i-1, right smaller
table[i][j] += table[i-1][k]*table[i-1][j-1-k];
                  // both i-1

In addition, if the number of nodes in the left subtree is smaller than the number of nodes in the left subtree, we can count the tree twice, as different tree can be constructed by swapping left and right subtree.

Total running time is O(K*N^2),with very favorable constant factor.

#include <cstdio>
#include <cstdlib>
#include <cassert>
#define MOD 9901
using namespace std;

int table[101][202],N,K,c;
int smalltrees[101][202];

FILE *fin=fopen("nocows.in","r");
FILE *fout=fopen("nocows.out","w");

int main() {
    fscanf (fin,"%d %d",&N,&K);
    table[1][1]=1;
    for (int i=2;i<=K;i++) {
        for (int j=1;j<=N;j+=2)
            for (int k=1;k<=j-1-k;k+=2) {
                if (k!=j-1-k) c=2; else c=1;    
                table[i][j]+=c*(
                        smalltrees[i-2][k]*table[i-1][j-1-k]  // left subtree smaller than i-1
                        +table[i-1][k]*smalltrees[i-2][j-1-k]  // right smaller
                        +table[i-1][k]*table[i-1][j-1-k]);// both i-1
                table[i][j]%=MOD;
            }
        for (int k=0;k<=N;k++) {          // we ensure that smalltrees[i-2][j] in
the next i
            smalltrees[i-1][k]+=table[i-1][k]+smalltrees[i-2][k]; // iteration
contains the number
            smalltrees[i-1][k]%=MOD;           // of trees smaller than i-1 and with
j nodes
        }
    }
    
    fprintf (fout,"%d\n",table[K][N]);
    return 0;
}