USACO: Cow Pedigrees

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#tree #output #structure #properties #integer #each

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本文介绍了一种通过递归算法计算特定规模和高度的牛谱系结构数量的方法，并提供了C++实现代码。该算法利用了动态规划思想，通过预先计算不同规模和高度的谱系结构数量，最终求得符合特定节点数和高度的所有可能谱系结构数量模9901的结果。

Cow Pedigrees

Silviu Ganceanu -- 2003

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      @   @
        / /                     / /
       @   @                   @   @

又一道递归思路的题目，而且你要学到的东西是，递归不一定要写递归函数来实现，实际上，我这里的报告中，多数递归都是类似这样的数组实现

/* ID: geterns1 PROG: nocows LANG: C++ */ #include <iostream> #include <fstream> #include <cstring> using namespace std; const unsigned id[] = { 0, 1, 3, 7, 15, 31, 63, 127 }; int main(){ ifstream ifs("nocows.in"); ofstream ofs("nocows.out"); unsigned n, k; ifs >> n >> k; unsigned lowBound[100], upBound[100]; for (unsigned i = 1; i < 100; i ++){ lowBound[i] = 2 * i - 1; upBound[i] = n; if (i <= 7 && id[i] < n) upBound[i] = id[i]; } unsigned tree[400][100] = {}; //node: 1 height: 1 tree[1][1] = 1; for (unsigned i = 1; i <= 98; i ++){ for (unsigned j = lowBound[i]; j <= upBound[i]; j += 2){ // for (unsigned k = 1; k < i; k ++){ for (unsigned l = lowBound[k]; l <= upBound[k]; l += 2){ tree[j + l + 1][i + 1] += tree[j][i] * tree[l][k] * 2; tree[j + l + 1][i + 1] %= 9901; } } } for (unsigned j = lowBound[i]; j <= upBound[i]; j += 2){ tree[2 * j + 1][i + 1] += tree[j][i] * tree[j][i]; tree[2 * j + 1][i + 1] %= 9901; for (unsigned k = j + 2; k <= upBound[i]; k += 2){ tree[j + k + 1][i + 1] += tree[j][i] * tree[k][i] * 2; tree[j + k + 1][i + 1] %= 9901; } } } ofs << tree[n][k] << endl; ifs.close(); ofs.close(); return 0; }