Singer House CodeForces

大意: 一个$k$层完全二叉树, 每个节点向它祖先连边, 就得到一个$k$房子, 求$k$房子的所有简单路径数.

$DP$好题.

首先设$dp_{i,j}$表示$i$房子, 分出$j$条简单路径的方案数, 那么最终答案就为$dp_{i,1}$.

考虑两棵$i-1$房子转移到$i$房子的情况, 分四种情况.

• 两个子树间不与根节点连边, 那么$dp_{i,j+k}=sum dp_{i-1,j}dp_{i-1,k}$
• 两个子树只有一条路径与根节点连边, $dp_{i,j+k}=sum dp_{i-1,j}dp_{i-1,k} 2(j+k)$
• 两个子树有两条路径与根节点连边, $dp_{i,j+k-1}=sum dp_{i-1,j}dp_{i-1,k} (j+k)(j+k-1)$
• 两个子树间不与根节点连边, 根节点单独作为一条路径, $dp_{i,j+k+1}=sum dp_{i-1,j}dp_{i-1,k}$

#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 '
'
#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
", dp[n][1]);
}