Effect: a $ k $ layer binary tree, each node is connected to its ancestor side, $ k $ get a house, find the number of all simple paths $ k $ house.
$ DP $ good question.
First set $ dp_ {i, j} $ I $ $ represents the house, the number of programs separated J $ $ article simple path, then the final answer is $ dp_ {i, 1} $ .
Consider two $ i-1 $ house $ i $ transitions into the house, sub-four cases.
- Between two sub-trees are not connected with the side of the root, then the $ dp_ {i, j + k} = \ sum dp_ {i-1, j} dp_ {i-1, k} $
- Only two sub-tree root with even one path side, $ dp_ {i, j + k} = \ sum dp_ {i-1, j} dp_ {i-1, k} 2 (j + k) $
- Two subtree root node there are two paths with even edges, $ dp_ {i, j + k-1} = \ sum dp_ {i-1, j} dp_ {i-1, k} (j + k) (j + k-1) $
- Between two sub-trees are not connected with the side of the root, the root node as a path alone, $ dp_ {i, j + k + 1} = \ sum dp_ {i-1, j} dp_ {i-1, k} $
#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head
const int N = 510;
int n, dp[N][N];
void add(int &x, ll y) {x=(x+y)%P;}
int main() {
scanf("%d", &n);
dp[1][1] = dp[1][0] = 1;
REP(i,2,n) {
REP(j,0,n) if (dp[i-1][j]) {
REP(k,0,n-j) if (dp[i-1][k]) {
ll t = (ll)dp[i-1][j]*dp[i-1][k]%P;
add(dp[i][j+k],t);
add(dp[i][j+k],t*2*(j+k));
if (j+k) add(dp[i][j+k-1],t*(j+k)*(j+k-1));
add(dp[i][j+k+1],t);
}
}
}
printf("%d\n", dp[n][1]);
}