HDU 4359 Easy Tree DP?

Easy Tree DP?

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1487    Accepted Submission(s): 567

Problem Description

A Bear tree is a binary tree with such properties : each node has a value of 2⁰,2¹…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 10⁹+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

解题：哎。。连续训练了两个月，身累心更累。。不想说什么了，看这位大神的解说吧

他的钻头是可以突破天际的钻头

#include <bits/stdc++.h>
using namespace std;
const int maxn = 365;
typedef long long LL;
const LL mod = 1e9 + 7;
LL dp[maxn][maxn],c[maxn][maxn];
void init(){
    memset(dp,-1,sizeof dp);
    for(int i = 0; i < maxn; ++i){
        c[i][0] = c[i][i] = 1;
        for(int j = 1; j < i; ++j)
            c[i][j] = (c[i-1][j-1] + c[i-1][j])%mod;
    }
}
LL dfs(int n,int d){
    if(n == 1 && d >= 1) return 1;
    if(n == 1 || d == 0) return 0;
    if(dp[n][d] != -1) return dp[n][d];
    LL &ret = dp[n][d];
    ret = (dfs(n-1,d-1)*c[n][1]*2)%mod;
    for(int k = 1; k <= n-2; ++k)
        ret = (ret + (dfs(n-k-1,d-1)*dfs(k,d-1)%mod*c[n-2][k]%mod*c[n][1])%mod)%mod;
    return ret;
}
int main(){
    int kase,cs = 1,n,d;
    init();
    scanf("%d",&kase);
    while(kase--){
        scanf("%d%d",&n,&d);
        printf("Case #%d: %I64d
",cs++,(dfs(n,d) - dfs(n,d-1) + mod)%mod);
    }
    return 0;
}