UVA 10253 Series-Parallel Networks （树形dp）

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Series-Parallel Networks

Input: standard input

Output:  standard output

Time Limit: 5 seconds

Memory Limit: 32 MB

In this problem you are expected to count two-terminal series-parallel networks. These are electric networks considered topologically or geometrically, that is, without the electrical properties of the elements connected. One of the two terminals can be considered as the source and the other as the sink.

A two-terminal network will be considered series-parallel if it can be obtained iteratively in the following way:

q       A single edge is two-terminal series-parallel.

q       If G₁ and G₂ are two-terminal series-parallel, so is the network obtained by identifying the sources and sinks, respectively (parallel composition).

q       If G₁ and G₂ are two-terminal series-parallel, so is the network obtained by identifying the sink of G₁with the source of G₂ (series composition).

Note here that in a series-parallel network two nodes can be connected by multiple edges. Moreover, networks are regarded as equivalent, not only topologically, but also when interchange of elements in series brings them into congruence; otherwise stated, series interchange is an equivalence operation. For example, the following three networks are equivalent:

　　　　　

Similarly, parallel interchange is also an equivalence operation. For example, the following three networks are also equivalent:

Now, given a number N, you are expected to count the number of two-terminal series parallel networks containing exactly N edges. For example, for N = 4, there are exactly 10 series-parallel networks as shown below:

Input

Each line of the input file contains an integer N (1 £N£ 30) specifying the number of edges in the network.

A line containing a zero for N terminates the input and this input need not be considered.

Output

For each N in the input file print a line containing the number of two-terminal series-parallel networks that can be obtained using exactly N edges.

 

Sample Input

1

4

15

0

 

Sample Output

1

10

1399068

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(World Final Warm-up Contest, Problem Setter: Rezaul Alam Chowdhury)

这道题目想了好久，最终还是参考了题解。

大致意思就是给你n条边，问你恰好用n条边，能构成几种串并联网络。（串联的各个部分可以任意调换，并联在一起的各个部分也可以任意调换，若通过调换可得，则二者视为等效）

分析：将每个网络都看成一棵树，为每次串联或者并联创建一个结点，并且把串联/并联部分看作该结点的子树，则可以转化为树形dp。

dp[i][j]表示每棵子树叶子数目不超过i，一共有j片叶子的方案数。

f[i]=dp[i-1][i],则根据可重复组合的公式，在有k个恰好包含i片叶子的子树时，其方案数等于C（f[i]+k-1,k）;

dp[i][j]=∑(C(f[i]+k-1,k)*d[i-1][j-p*i])     k≥0,k*i<=j

另外注意处理好边界。

对于求这个组合数，想不出较好的方法，最终还是采用了刘汝佳在大白书上写的用double来做的方法(虽然我一度担心会因为double的精度问题会使得有所误差)。

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 using namespace std;
 typedef long long ll;
 ll dp[][];
 ll f[];
 ll C(ll n,int m)
 {
     double ret=;
     for(ll i=n+-m;i<=n;i++)
     {
         ret*=i;
     }
     for(int i=;i<=m;i++)ret/=i;
     return (ll)(ret+0.5);
 }
 int main()
 {
     ios::sync_with_stdio(false);
     int n=;
     f[]=;
     for(int i=;i<=n;i++){dp[][i]=;dp[i][]=;}
     for(int i=;i<=n;i++)dp[i][]=;
     for(int i=;i<=n;i++)
     {
         for(int j=;j<=n;j++)
         {
             dp[i][j]=;
             for(int k=;i*k<=j;k++)
             {
                 dp[i][j]+=C(f[i]+k-,k)*dp[i-][j-i*k];
             }
         }
         f[i+]=dp[i][i+];
     }
     for(int i=;i<=n;i++)f[i]*=2LL;
     while(cin>>n&&n)
     {
         cout<<f[n]<<endl;
     }
 }