bzoj3769

Posted 2020-10-06 123456

树形dp

%%%popoqqq

设dp[i][j]表示当前i个节点的树，深度小于等于j的树的个数

那么dp[i][j] = sigma(dp[k][j-1]*dp[n-k-1][j-1]) 比较好理解 然后记忆化搜索就行了

这个dp状态感觉挺巧妙的，以前没见过

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
typedef long long ll;
const int N = 610, mod = 1000000007;
int n, h, T;
ll dp[N][N];
ll dfs(int n, int h)
{
    if(n < 0 || h < 0) return 0;
    if(n == 0) return 1;
    if(h == 0) return n == 1;
    if(dp[n][h] != -1) return dp[n][h];
    ll ret = 0;
    for(int i = 0; i < n; ++i) ret = (ret + dfs(i, h - 1) * dfs(n - i - 1, h - 1)) % mod;
    return dp[n][h] = ret;
}
int main()
{
    memset(dp, -1, sizeof(dp));
    for(cin >> T; T; --T)
    {
        cin >> n >> h;
        cout << ((dfs(n, h) - dfs(n, h - 1)) % mod + mod) % mod << endl;
    }
    return 0;
}

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