# USACO 2.3 Cow Pedigrees 奶牛家谱 解题报告

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)
2

OUTPUT DETAILS
Two possible pedigrees have 5 nodes and height equal to 3:
@                   @
/ \                 / \
@   @      and      @   @
/ \                     / \
@   @                   @   @

USACO_2.3-2:Cow Pedigrees奶牛家谱

Time Limit:1000MS  Memory Limit:65536K
Total Submit:1 Accepted:1

Description

Input

PROGRAM NAME: nocows

Output

Sample Input

SAMPLE INPUT (file nocows.in)

5 3

Sample Output

SAMPLE OUTPUT (file nocows.out)

2

======================== 华丽的分割线 ========================
题目是看懂了,, 意思就是说用N个节点构造一个高度为K的二叉树, 且每个节点的度不能为1(即只能要么有两个儿子, 要么就没有没有孩子.), 题目我是理解了, 也清楚得很是用DP.. 但是不太会做.. DP没学好.. (自卑中)
终于吧DP返程看懂了.~!~!~! 狂High中..
f[i][j] 代表用i各节点构成最多j层的二叉树有多少种情况, 那么
f[i][j] = ∑(f[m][j-1] * f[i-1-m][j-1])(m = 1, 2, 3 … i – 1)
m是左子树的节点个数,, 那么i – m 就是除了左子树节点个数之外的节点个数, 再除根节点, 即 i – m – 1就是右子树的节点个数..
代码能有两个优化的地方(我能够想到的只有这两个),
第一个就是在DP方程中f[i][j] 和 f[m][j] 都一定是奇数, 因为如果是偶数的话就不能够构成题目所要求的二叉树了.
第二个就是数据是能够对折的. 这个我晚点尝试一下.. 现在先把没折半的发一下吧.

/*
PROG: nocows
ID: zqy11001
LANG: C
*/
#include <stdio.h>

int f[200][100];

int main(void)
{
int n, m;
int i, j, k;
freopen(nocows.in, r, stdin);
freopen(nocows.out, w, stdout);
scanf(%d%d, &n, &m);
for(j = 1; j <= m; j++){
f[1][j] = 1;
}
for(j = 2; j <= m; j++){
for(i = 1; i <= n; i += 2){
for(k = 1; k <= i - 2; k++){
f[i][j] += f[k][j - 1] * f[i - k - 1][j - 1];
f[i][j] %= 9901;
}
}
}
printf(%d\\n, (9901 + f[n][m] - f[n][m - 1]) % 9901);
return 0;
}