Codeforces Beta Round #9 (Div. 2 Only) D. How many trees? dp
2017-03-23 15:23
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题目链接:http://codeforces.com/contest/9/problem/D
题意:
问你一个二叉树,有n个节点,深度大于等于h的一共有多少种
这个二叉树满足左儿子比自己小,右儿子比自己大的特性。
解法:
dp,dp[i][j]表示当前用了i个节点,高度小于等于j的方案数
dp[i][j] = sigma(dp[k][j-1]*dp[i-k-1][j-1])
枚举左子树和右子树
然后莽一波就好了
题意:
问你一个二叉树,有n个节点,深度大于等于h的一共有多少种
这个二叉树满足左儿子比自己小,右儿子比自己大的特性。
解法:
dp,dp[i][j]表示当前用了i个节点,高度小于等于j的方案数
dp[i][j] = sigma(dp[k][j-1]*dp[i-k-1][j-1])
枚举左子树和右子树
然后莽一波就好了
//CF 9D #include <bits/stdc++.h> using namespace std; const int maxn = 37; long long dp[maxn][maxn]; //dp[i][j]表示当前用了i个节点高度小于等于j的方案数 int main() { int n, h; scanf("%d%d", &n, &h); for(int i = 1; i <= n; i++){ dp[0][i-1] = 1; for(int j = 1; j <= n; j++) for(int k = 0; k < j; k++) dp[j][i] += dp[k][i-1]*dp[j-k-1][i-1]; } cout << dp - dp [h-1] << endl; return 0; }
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