题目链接：uva 10253 - Series-Parallel Networks

题目大意：就是有n条线，通过并联或者是串联，形成一个整体，问说有n条线，可以组成多少种。

解题思路：大白书上的例题，解法还真是高端.dp[i][j]表示说每个树德叶子节点不大于i，一共有j个叶子。f[i]=dp[i-1][i]，注意n为1的时候。

#include 
   
     #include 
    
      typedef long long ll; const int N = 35; ll f[N], dp[N][N]; ll C(ll n, ll m) { double ans = 1; for (ll i = 0; i < m; i++) ans *= n - i; for (ll i = 1; i <= m; i++) ans /= i; return (ll)(ans+0.5); } void init () { f[1] = 1; int n = 30; memset(dp, 0, sizeof(dp)); for (int i = 0; i <= n; i++) dp[i][0] = 1; for (int i = 1; i <= n; i++) { dp[i][1] = 1; dp[0][i] = 0; } for (int i = 1; i <= n; i++) { for (int j = 2; j <= n; j++) { // dp[i][j] = 0; for (int p = 0; p * i <= j; p++) dp[i][j] += dp[i-1][j-i*p] * C(f[i]+p-1, p); } f[i+1] = dp[i][i+1]; } } int main () { init (); int n; while (scanf("%d", &n) == 1 && n) { printf("%lld\n", n == 1   1 : 2 * f[n]); } return 0; }