HDOJ 4359 Easy Tree DP? DP

Easy Tree DP?

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1482    Accepted Submission(s): 565

Problem Description

A Bear tree is a binary tree with such properties : each node has a value of 2 ⁰,2 ¹…2 ^(N-1)(each number used only once),and for each node ,its left subtree’s elements’ sum<its right subtree’s elements’ sum(if the node hasn’t left/right subtree ,this limitation is invalid).
You need to calculate how many Bear trees with N nodes and exactly D deeps.

 

Input

First a integer T(T<=5000),then T lines follow ,every line has two positive integer N,D.(1<=D<=N<=360).

 

Output

For each test case, print "Case #t:" first, in which t is the number of the test case starting from 1 and the number of Bear tree.(mod 10 ⁹+7)

 

Sample Input

   
   2
2 2
4 3
  
  

 

Sample Output

   
   Case #1: 4
Case #2: 72
  
  

 

Author

smxrwzdx@UESTC_Brightroar

 

Source

2012 Multi-University Training Contest 6

 

/* ***********************************************
Author        :CKboss
Created Time  :2015年09月04日 星期五 16时40分18秒
File Name     :1009.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

typedef long long int LL;

const int maxn=400;
const LL mod=1e9+7;

LL dp[maxn][maxn];
LL C[maxn][maxn];

void init()
{
	for(int i=1;i<maxn;i++) C[i][0]=C[i][i]=1LL;
	for(int i=1;i<maxn;i++)
	{
		for(int j=1;j<i;j++)
		{
			C[i][j]=(C[i-1][j]+C[i-1][j-1])%mod;
		}
	}

	for(int i=1;i<=360;i++) dp[1][i]=1;
	for(int i=2;i<=360;i++) dp[2][i]=4;

	for(int i=3;i<=360;i++)
	{
		for(int j=1;j<=360;j++)
		{
			dp[i][j]=(dp[i][j]+dp[i-1][j-1]*2*i)%mod;
			for(int left=1;left<=i-2;left++)
			{
				int right=i-1-left;
				dp[i][j]=(dp[i][j]+dp[left][j-1]*dp[right][j-1]%mod*i%mod*C[i-2][left]%mod)%mod;
			}
		}
	}
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    
	init();
	int T_T,cas=1,n,d;
	scanf("%d",&T_T);
	while(T_T--)
	{
		scanf("%d%d",&n,&d);
		printf("Case #%d: %lld\n",cas++,(dp[n][d]-dp[n][d-1]+mod)%mod);
	}
	return 0;
}